CH. 4: Stoichiometry of Chemical Reaction

The oxidation number on the carbon in CH₃OH is

-2; Oxygen has an oxidation number of -2 when not in its elemental state. Hydrogen has an oxidation number of +1 when not in its elemental state. Multiply the number of hydrogen atoms by its oxidation number. Add the oxidation numbers of all atoms together and set equal to the charge on the molecule, which is zero. The oxidation number of carbon can then be calculated. C+(4x(+1))+-2=0 C=-2

How many moles of precipitate will be formed when 61.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂ in the following chemical reaction? 2 NaI (aq) + Pb(NO₃)₂ (aq) → PbI₂ (s) + 2 NaNO₃ (aq)

0.0610mol PbI₂; Step 1) The volume and molarity are provided for each reactant and the amount of moles of each can be calculated: 61.0mL(2.00mol NaI/1000mL)=0.122mol NaI 100mL(0.900mol Pb(NO₃)₂/1000mL)= 0.090mol Pb(NO₃)₂ Step 2) The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur. The balanced reaction provides the conversion factor: 0.122mol NI(1mol Pb(NO₃)₂/2mol NaI)= 0.061molPb(NO₃)₂ 0.090mol Pb(NO₃)₂(2mol NaI/1mol Pb(NO₃)₂= 0.180mol NaI Step 3) NaI is the limiting reactant because there isn't enough of it to completely react with the amount of Pb(NO₃)₂. The amount of products can be found using conversion factors found in the balanced reaction: 0.122mol NaI(1mol PbI₂/2mol NaI)= 0.061mol PbI₂

How many moles of HNO₃ will be produced from the reaction of 41.5 g of NO₂ with excess water in the following chemical reaction? 3 NO₂(g) + H₂O (l) → 2 HNO₃(g) + NO(g)

0.601 mol HNO₃; The number of moles of substances in a reaction can be determined given a balanced equation and an amount of one substance. The number of grams of a substance can be converted to moles using the molar mass of the substance. 45.0g NO₂ (1 mol NO₂)/46.01g NO₂) (2 mol HNO₃/3 mol NO₂) = 0.601 mol

Determine the number of grams of HCl that can react with 0.750 g of Al(OH)₃ according to the following reaction: Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3 H₂O(aq)

0.750g Al(OH)₃ (1 mol Al(OH)₃/78.00g Al(OH)₃) (3 mol HCl/1 mol Al(OH)₃) (36.46g HCl/1 mol HCl) = 1.05g HCl

How many molecules of SO₃ can be formed from 0.12 moles of O₂ (assuming excess SO₂) from the following UNBALANCED equation? SO₂(g) + O₂(g) → SO₃(g)

1.4 × 10²³ molecules of SO₃. 0.12 mol O₂ (2 mol SO₃/1 mol O₂) (6.022x10²³ molecules SO₃/1 mol SO₃) = 1.4x10²³ molecules.

What volume (in mL) of 0.4000 M HCl is required to neutralize 50.00 mL of 0.8000 M NaOH?

100.0mL HCl; Step 1)The balanced reaction is HCl + NaOH → NaCl + H₂O; Begin by converting the milliliters of NaOH to liters: 50.00mL NaOH(1L/1000mL)= 0.05000L NaOH Step 2) Stoichiometry can be used to determine the volume of HCl required. The balanced reaction provides the mole ratio conversion factor: 0.05000L NaOH(0.800mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.4000mol HCl)=0.1L HCl Step 3) Next, convert the liters of HCl to milliliters: 0.1L HCl(1000mL/1L)=100.0mL HCl

Complete and balance the following redox reaction in acidic solution: Mn²⁺(aq) + BiO₃⁻(aq) → MnO₄⁻(aq) + Bi³⁺(aq)

14 H⁺(aq) + 2 Mn²⁺(aq) + 5 BiO₃⁻(aq) → 2 MnO₄⁻(aq) + 5 Bi³⁺(aq) + 7 H₂O(l)

How many grams of carbon dioxide would be required to react with 21.1 g of LiOH in the following chemical reaction? 2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

19.39g; The number of moles of substances in a reaction can be determined given a balanced equation and an amount of one substance. The number of grams of a substance can be converted to moles using the molar mass of the substance and vice versa. 21.1g LiOH(1 mol LiOH/23.95g LiOH) (1 mol CO₂/2 mol LiOH) (44.01g CO₂/1 mol CO₂)

Write a balanced chemical equation based on the following description: aqueous silver nitrate reacts with aqueous nickel(II) chloride to produce solid silver chloride and aqueous nickel(II) nitrate

2 AgNO₃(aq) + NiCl₂(aq) → 2 AgCl(s) + Ni(NO₃)₂(aq)

Write a balanced chemical equation based on the following description: solid chromium reacts with solid iodine to form solid chromium(III) iodide.

2 Cr(s) + 3 I₂(s) → 2 CrI₃(s)

How many liters of a 0.209M KI solution is needed to completely react with 2.43 g of Cu(NO₃)₂ according to the balanced chemical equation: 2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + 4KNO₃(aq)

2.43g Cu(NO₃)₂ (1mol Cu(NO₃)₂/187.57g Cu(NO₃)₂) (4mol KI/2mol Cu(NO₃)₂) (1L KI/0.209mol KI) = 0.124L KI

How many grams of NO can be formed from 2.26 moles of NO₂ in the reaction below? 3 NO₂ (g) + H₂O (g) → 2 HNO₃ (g) + NO (g)

22.6 g; The number of moles of substances in a reaction can be determined given a balanced equation and an amount of one substance. Then the number of grams can be found using the molar mass of the substance. 4.96 mol NO₂ (1 mol NO/3 mol NO₂) (30.01g NO/1 mol NO) = 49.62g NO

Write the balanced NET ionic equation for the reaction when CaBr₂ and Na₃PO₄ are mixed in aqueous solution. If no reaction occurs, write only NR.

3Ca²⁺(aq) + 2PO₄³⁻(aq) → Ca₃(PO₄)₂(s)

The oxidation number on the nitrogen in NO₃⁻ is

5; Oxygen has an oxidation number of -2 when not in its elemental state. Multiply the number of oxygen atoms by its oxidation number. Add the oxidation numbers of all atoms together and set equal to the charge on the ion, which is -1. The oxidation number of nitrogen can then be calculated. N+(3x(-2))=-1 N=+5

The oxidation number on the phosphorus in Na₃PO₄ is

5; Oxygen has an oxidation number of -2 when not in its elemental state. Multiply the number of oxygen atoms by its oxidation number. Sodium has an oxidation number of +1 when not in its elemental state. Multiply the number of sodium atoms by its oxidation number. Add the oxidation numbers of all atoms together and set equal to the charge on the molecule, which is zero. The oxidation number of phosphorus can then be calculated. P=(4x(-2))+(3x(+1))=0 P=+5

What mass in grams of NaN₃ is required to produce 50.2L of N₂ gas (density=1.25g/L) according to the balanced chemical reaction: 2NaN₃(s) → 2Na(s) + 3N₂(g)

97.1g NaN₃

In the following chemical reaction, which element is the reducing agent? 2 Ag(s) + 2Cl⁻(aq) + 2 H₂O(l) → 2 AgCl(s) + H₂(g) + 2 OH⁻(aq) A) Ag B) Cl C) H D) O

A) Ag; Reducing agents lose electrons and are oxidized. In this reaction, Ag loses one electron to become Ag⁺.

A 15.79 g sample of a hydrate of an iron(II) sulfate compound was heated, without decomposing the sulfate to drive off the water. The mass was reduced to 8.63 g . What is the formula of the hydrate? A) FeSO₄･7 H₂O B) Fe₂(SO₄)₃･7 H₂O C) Fe₂(SO₄)₃･18 H₂O D) FeSO₄･16 H₂O

A) FeSO₄･7 H₂O The formula of the hydrate can be determined from the masses of the sample before and after the water was removed. 15.79g FeSO₄∙xH₂O-8.63g FeSO₄= 7.16g H₂O 8.13g FeSO₄(1mol FeSO₄/151.91g FeSO4)= 0.0568mol FeSO₄ 7.16g H₂O(1mole H₂O/18.02g H₂O)=0.397mol H₂O 0.397mol H₂O/0.0568mol FeSO₄=7mol H₂O/1mol FeSO₄

In which of the following species does sulfur have the lowest oxidation number? A) S²⁻ B) S₈ C) SO₃²⁻ D) SO₄²⁻ E) S₂O₃²⁻

A) S²⁻; The oxidation number of an ion is the charge on the ion. S²⁻ has an oxidation number of -2. S₈ has an oxidation number of 0, and a positive oxidation number in the other species, so S²⁻ has the lowest oxidation number.

What type of reaction is represented by the following equation: 2Na(s) + Cl₂(g) → 2NaCl(s) A) combination B) decomposition C) double displacement D) single displacement

A) combination; This is a combination reaction, involves two compounds reacting to form one new compound. This type of reaction is also known as a synthesis reaction.

Balance the following chemical equation (if necessary): Ag₂CO₃(s) ⇌ Ag⁺(aq) + CO₃²⁻(aq)

Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq)

In the following reaction, which element in what species is reduced? Zn (s) + 2 HCl (aq) → ZnCl₂ (aq) + H₂ (g) A) Zn B) H in HCl C) Cl in HCl D) This is not an oxidation/reduction type of reaction.

B) H in HCl; Oxidizing agents gain electrons and are reduced. H⁺ in HCl gains an electron when it becomes H₂.

What is the oxidizing agent in the reaction Zn + 2 H⁺ → Zn²⁺ + H₂? A) Zn B) H⁺ C) Zn²⁺ D) H₂ E) This is not a redox reaction.

B) H⁺; Zn increases in oxidation number from 0 to +2 and H⁺ decreases in oxidation number from +1 to 0. Therefore Zn is being oxidized, and H⁺ is the oxidizing agent.

Write the balanced molecular chemical equation for the reaction in aqueous solution for barium nitrate and lithium sulfate. If no reaction occurs, write NR.

Ba(NO₃)₂(aq) + Li₂SO₄(aq) → 2 LiNO₃(aq) + BaSO₄(s)

Write a balanced chemical equation based on the following description: solid barium carbonate decomposes into solid barium oxide and carbon dioxide gas when heated

BaCO₃(s) → BaO(s) + CO₂(g)

If you have 1 mol Xe and 1 mol F₂, how many moles of XeF₄ can you create in the following chemical reaction? Xe (g) + 2 F₂ (g) → XeF₄ (g) A) 2 mol B) 1 mol C) 0.5 mol D) 0.25 mol

C) 0.5 mol; Calculate the amount of XeF₄ that could be made from the given amount of each reactant, if it were not limiting: 1mol Xe (1mol XeF₄/1mol Xe) = 1mol XeF₄ 1mol F₂ (1mol XeF₄/2mol F₂) = 0.5 mol XeF₄ Therefore F₂ is the limiting reactant, and 0.5 mol of XeF₄ will be produced.

A 10.68 g sample of compound that contains only carbon, hydrogen, and oxygen was analyzed through combustion analysis and yielded 16.01 g of CO₂ and 4.37 g of H₂O. What is this compound's empirical formula? A) C₄H₁₃O₁₅ B) CH₁₃O C) C₃H₄O₃ D) C₃H₂O₃

C) C₃H₄O₃; An empirical formula shows the ratio of each element to one another in the formula unit. The mass of carbon dioxide can be used to determine the mass of carbon that was reacted in the sample via a proportion. This can be converted to moles using the molar mass. The same can be repeated for the mass of water to determine the moles of hydrogen that were reacted in the sample. Then simplify the ratio between between them such that both are whole numbers: 1) 44.01g CO₂/12.01g C=16.01g CO₂/x 2) (12.01g Cx16.01g CO₂)/44.01gCO₂ =4.27gC(1molC/12.01gC)=0.364molC 3) 18.02gH₂O/2.02gH=4.37gH₂O/x 4)(2.02gHx4.37gH₂O)/18.02gH₂O= 0.490gH(1molH/1.01gH)=0.485molH 5)0.485molH/0.364molC=1.3molH/1molC(3/3)=4molH/3molC

Write the balanced NET ionic equation for the reaction when copper(II) sulfate and ammonium hydroxide are mixed in aqueous solution. If no reaction occurs, write only NR.

Cu²⁺(aq) + 2 OH⁻(aq) → Cu(OH)₂(s)

Write a balanced chemical equation based on the following description: solid C₄H₁₀O is burned with oxygen gas to produce gaseous carbon dioxide and water vapor

C₄H₁₀O(s) + 6 O₂(g) → 4 CO₂(g) + 5 H₂O(g)

Which of the following reactions is a precipitation reaction? A) HNO₂(aq) + NaOH(aq) → NaNO₂(aq) + H₂O(l) B) 2C₂H₅OH(l) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) C) 2H₂O₂(l) → 2H₂O(l) + O₂(g) D) FeSO₄(aq) + K₂S(aq) → FeS(s) + K₂SO₄(aq)

D) FeSO₄(aq) + K₂S(aq) → FeS(s) + K₂SO₄(aq); FeSO₄(aq) + K₂S(aq) → FeS(s) + K₂SO₄(aq), is a precipitation reaction involves the formation of an insoluble solid and an aqueous solution.

Which choice correctly identifies the oxidation number (O.N.) for the given species? A) Mn²⁺, O.N. = 0 B) Ca, O.N. = +2 C) F₂, O.N. = -1 D) Se²⁻, O.N. -2 E) P₄, O.N. = -1

D) Se²⁻, O.N. -2; Only Se²⁻, O.N. -2 is Correct! The oxidation number of an ion is the charge on the ion.

Which one of the following would form a precipitate when mixed with LiOH? A) KNO₃ B) NH₄Cl C) Ca(C₂H₃O₂) D) ZnBr₃ E) None of these would precipitate

D) ZnBr₃; Zinc hydroxide is insoluble in water. Although most hydroxides are insoluble in water, potassium, ammonium and calcium hydroxides are exceptions.

The reaction of an alkane with oxygen to produce carbon dioxide and water is known as what? A) substitution B) displacement C) precipitation D) combustion E) double replacement

D) combustion; Combustion comes from the Latin comburere, "to burn up, consume."

Consider the following reaction: Mg²⁺(aq) + Cu(s) → Cu²⁺(aq) + Mg(s). In this reaction, Mg²⁺(aq) is: A) Oxidized B) Reduced C) A reducing agent D) An oxidizing agent E) Both (B) and (D)

E) Both (B) and (D); Oxidizing agents gain electrons and are reduced. Reducing agents lose electrons and are oxidized. In this reaction, Mg²⁺ gains two electrons when it becomes Mg.

What are the reactants in the following equation: HCl(aq) + NaHCO₃(aq)→ CO₂(g) + H₂O(l) + NaCl(aq)

HCl, NaHCO₃

Complete the balanced molecular chemical equation for the reaction below. If no reaction occurs, write NR after the reaction arrow. KBr(aq) + CaI₂(aq) →

KBr(aq) + CaI₂(aq) → NR

Complete the balanced molecular chemical equation for the reaction below. If no reaction occurs, write NR after the reaction arrow. K₂SO₄(aq) + FeCl₃(aq) →

K₂SO₄(aq) + FeCl₃(aq) → NR

Write the balanced molecular chemical equation for the reaction in aqueous solution for magnesium chloride and silver nitrate. If no reaction occurs, write NR.

MgCl₂(aq) + 2 AgNO₃(aq) → Mg(NO₃)₂(aq) + 2 AgCl(s)

Write the balanced NET ionic equation for the reaction when magnesium iodide and sodium carbonate are mixed in aqueous solution. If no reaction occurs, write only NR.

Mg²⁺(aq) + CO₃²⁻(aq) → MgCO₃(s)

Complete and balance the following half-reaction in acidic solution: MnO₄⁻(aq) → MnO₂(s)

MnO₄⁻(aq) + 4 H⁺(aq) + 3 e⁻→ MnO₂(s) + 2 H₂O(l)

Complete and balance the following half-reaction in basic solution: O₂(g) → H₂O(l)

O₂(g) + 2 H₂O(l) + 4 e⁻ → 4 OH⁻(aq)

Complete and balance the following half-reaction in acidic solution: Sn²⁺(aq) → Sn⁴⁺(aq)

Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻

If 20.5 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

0.090mol O₂ excess; Step 1) The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur: 20.5g NO(1mol NO/30.01g NO)= 0.683mol NO 0.683mol NO(1mol O₂/2mol NO)= 0.34155mol O₂ 13.8g O₂(1mol O₂/32.00g O₂) = 0.431mol O₂ Step 2) NO is the limiting reactant because there isn't enough of it to completely react with the amount of O₂. The amount of O₂ used in the reaction can be found using the amount of the limiting reactant NO and mole to mole ratios. Excess O₂ is then calculated by taking the difference between the starting amount of O₂ and the amount of O₂ used in the reaction. 0.683mol NO(1mol O₂/2mol NO) = 0.34155mol O₂ 0.431mol O₂ - 0.34155mol O₂ = 0.0894mol O₂ excess

A 25.0 mL solution of HNO₃ is neutralized with 11.9 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?

0.238M HNO₃; The balanced reaction is 2HNO₃ + Ba(OH)₂→Ba(NO₃)₂ + 2H₂O. Step 1)Begin by converting the milliliters of Ba(OH)₂ to liters: 11.9mL Ba(OH)₂(1L/1000mL)=0.0119L Ba(OH)₂ Step 2)Stoichiometry using the balanced reaction provides the conversion factor to determine the moles of HNO₃: 0.0119L Ba(OH)₂(0.250mol Ba(OH)₂/1L)(2mol HNO₃/1mol Ba(OH)₂)= 0.00595mol HNO₃ Step 3)The number of moles of HNO₃ can then be divided by the number of liters used to find the molarity: 25.0mL HNO₃(1L/1000mL)=0.0250L HNO₃ 0.00595mol HNO₃/0.0250L = 0.238M HNO₃

Determine the number of molecules of Cr₂O₃ that form when 1.34 × 10³g of oxygen completely reacts according to the following equation: 4Cr(s) + 3O₂(g) → 2Cr₂O₃(s)

1.68 × 10²⁵ molecules of Cr₂O₃

Balance the following chemical equation (if necessary): C₆H₆(l) + O₂(g) → H₂O(g) + CO₂(g)

2 C₆H₆(l) + 15 O₂(g) → 6 H₂O(g) + 12 CO₂(g)

How many moles of C are formed upon complete reaction of 2 mol of B according to the generic chemical reaction: A + 2B → C

2 mol B (1 mol C/2 mol B) = 1 mol C

If 2.05 moles of H₂ and 1.55 moles of O₂ react how many moles of H₂O can be produced in the reaction below? 2 H₂(g) + O₂(g) → 2 H₂O(g)

2.05mol H₂O; The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur: 2.05mol H₂(1mol O₂/2mol H₂)=1.025mol O₂ 1.55mol O₂(2molH₂/1mol O₂)=3.10mol H₂; H₂ is the limiting reactant because there isn't enough of it to completely react with the amount of O₂. The amount of other products and reactants can be found using the amount of H₂ and mole to mole ratios. 2.05mol H₂(2mol H₂O/2mol H₂)= 2.05mol H₂O

What volume (in mL) of 0.300 M HCl would be required to completely react with 5.90 g of Al in the following chemical reaction? 2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

2187mL HCl; Step 1)The number of moles can be calculated using the molar mass and the balanced equation provides the conversion factor: 5.90g Al(1mol Al/26.98g Al)= 0.21868mol Al 0.21868mol(6mol HCl/2mol Al)= 0.65605mol HCl Step 2) A proportion of moles and volume can be used to find the volume needed: 0.300mol HCl/1000mL=0.65605molHCl/x x=0.65605mol HCl(1000mL)/0.300mol HCl=2186.83mL

Write the balanced COMPLETE ionic equation for the reaction when silver acetate and magnesium chloride are mixed in aqueous solution. If no reaction occurs, write only NR.

2Ag⁺(aq) + 2C₂H₃O₂⁻(aq) + Mg²⁺(aq) + 2Cl⁻(aq) → 2AgCl(s) + 2C₂H₃O₂⁻(aq) + Mg²⁺(aq)

Balance the following chemical equation (if necessary): Al(s) + HCI(aq) → H₂(g) + AlCl₃(aq)

2Al(s) + 6HCI(aq) → 3H₂(g) + 2AlCl₃(aq)

Balance the following chemical equation (if necessary): Cl⁻(aq) + Pb²⁺(aq) ⇌ PbCl₂(s)

2Cl⁻(aq) + Pb²⁺(aq) ⇌ PbCl₂(s)

Balance the following chemical equation (if necessary): H₂O(l) → H₂(g) + O₂(g)

2H₂O(l) → 2H₂(g) + O₂(g)

Balance the following chemical equation (if necessary): KOH(aq) + H₂S(aq) → H₂O(l) + K₂S(aq)

2KOH(aq) + H₂S(aq) → 2H₂O(l) + K₂S(aq)

Complete the balanced molecular chemical equation for the reaction below. If no reaction occurs, write NR after the reaction arrow. KOH(aq) + AlCl₃(aq)

3 KOH(aq) + AlCl₃(aq) → Al(OH)₃(s) + 3 KCl(aq)

Complete and balance the following redox reaction in basic solution: O₂(g) + Al(s) → H₂O₂(aq) + AlO₂⁻(aq)

3 O₂(g) + 2 Al(s) + 2 OH⁻(aq) + 2 H₂O(l) → 3 H₂O₂(aq) + 2 AlO₂⁻(aq)

If 50.0 g of H₂ and 51.0 g of O₂ react, how many moles of H₂O can be produced in the reaction below? 2 H₂(g) + O₂(g) → 2 H₂O(g)

3.188mol H₂O; The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur. The given mass can be converted to moles using molar mass: 50.00g H₂(1mol H₂/2.02g H₂) = 24.8mol H₂ 24.8mol H₂(1mol O₂/2mol H₂)= 12.4mol O₂ 51.0g O₂(1mol O₂/32.00g O₂)= 1.59375mol O₂ 1.59375mol O₂(2mol H₂ /1mol O₂)= 3.1875mol H₂ O₂ is the limiting reactant because there isn't enough of it to completely react with the amount of H₂. The amount of other products and reactants can be found using the amount of O₂ and mole to mole ratios. 1.59375mol O₂(2mol H₂O/1mol O₂)=3.1875mol H₂O

If 25.0 g of NH₃ and 49.1 g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

36.8g NO; Step 1) the grams of each substance need converted into moles using their molar masses: 25.0g NH₃(1mol NH₃/17.04g NH₃)= 1.467mol NH₃ 49.1g O₂(1mol O₂/32.00g O₂)=1.534mol O₂ Step 2) The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur: 1.467mol NH₃(5mol O₂/4mol NH₃)= 1.834mol O₂ 1.534mol O₂(4mol NH₃/5mol O₂)= 1.2275mol NH₃ Step 3) O₂ is the limiting reactant because there isn't enough of it to completely react with the amount of NH₃ present. The amount of other products can be found using the amount of O₂ and mole to mole ratios. The moles of NO can be converted into grams using the molar mass of NO: 1.534mol O₂(4mol NO/5mol O₂) = 1.22275mol NH₃ 1.2275mol NO(30.01g NO/1mol NO)=36.837g NO

How many grams of tin would be required to completely react with 1.75 L of 0.750 M HBr in the following chemical reaction? Sn(s) + 4 HBr(aq) → SnBr₄ (aq) + 2 H₂ (g)

39.0g Sn; The number of moles can be calculated using the given the molarity and volume of HBr. The balanced equation and molar mass provide the conversion factors: 1.75L (0.750mol HBr/1L) = 1.3125mol HBr 1.3125mol HBr(1mol Sn/4mol HBr)=0.328125mol Sn 0.328125mol Sn(118.71g Sn/1mol Sn)= 38.95gSn

Write the balanced COMPLETE ionic equation for the reaction when copper(I) nitrate and sodium phosphate are mixed in aqueous solution. If no reaction occurs, write only NR.

3Cu⁺(aq) + 3NO₃⁻(aq) + 3Na⁺(aq) + PO₄³⁻(aq) → Cu₃PO₄(s) + 3NO₃⁻(aq) + 3Na⁺(aq)

Balance the following chemical equation (if necessary): NH₃(g) + O₂(g) → NO(g) + H₂O(g)

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

According to the balanced reaction below, calculate the moles of NH₃ that form when 4.2 mol of N₂H₄ completely reacts 3 N₂H₄(l) → 4 NH₃(g) + N₂(g)

4.2 mol N₂H₄ (4 mol NH₃/3 mol N₂H₄) = 5.6 mol NH₃

How many moles of NH₃ can be produced from 2.36 moles of nitrogen in the following reaction: N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

4.72 mol. The number of moles of substances in a reaction can be determined given a balanced equation and an amount of one substance. 2.36 mol N₂ (3 mol NH₃/1 mol N₂) = 4.72 mol NH₃

What is the oxidation number of S in the compound Na₂SO₃?

4; O always has an oxidation number of -2 for compounds other than peroxides, and Na is always +1 when in a compound. Therefore in Na₂S₂O₅, S must oxidation number of +4.

The oxidation number on the nitrogen in NO₂ is

4; Oxygen has an oxidation number of -2 when not in its elemental state. Multiply the number of oxygen by its oxidation number. Add the oxidation numbers of all atoms together and set equal to the charge on the molecule, which is zero. The oxidation number of nitrogen can then be calculated. For NO₂: N+(2×(−2))=0 N+(2×(−2))=0N=+4

How many moles of AgI will be formed when 75.0mL of 0.300M AgNO₃ is completely reacted according to the balanced chemical reaction: 2AgNO₃(aq) + CaI₂(aq) → 2AgI(s) + Ca(NO₃)₂(aq)

75.0mL AgNO₃ (1L AgNO₃/1000mL AgNO₃) (0.300 mol AgNO₃/1L AgNO₃) (2mol AgI/2mol AgNO₃) = 0.0225mol AgI

When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 1.50 g of copper was heated with excess sulfur to yield 1.72 g copper(I) sulfide. What is the percent yield?

91.6%; Step 1) The mass of copper needs converted to moles using its molar mass: 1.50g Cu(1mol Cu/63.55g Cu)= 0.0236035mol Cu Step 2) The amount of other products can be found using the amount of Cu and mole to mole ratios. This is equal to the theoretical yield: 0.0236035mol Cu(1mol Cu₂S/2mol Cu)= 0.01180175 mol Cu₂S Step 3) The grams of the actual yield can be converted to moles using the molar mass: 1.72g Cu₂S(1mol Cu₂S/159.17g Cu₂S)= 0.010806056mol Cu₂S Step 4) The actual yield can be divided by the theoretical yield to get the percent yield: 0.010806056mol Cu₂S/0.01180175mol Cu₂S (100)= 91.56%

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 1.92 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what is the percent yield?

91.7%; Step 1) The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur: 1.92mol Mg(1mol I₂/1mol Mg)= 1.92mol I₂ 3.56mol I₂(1mol Mg/1mol I₂)= 3.56mol Mg Step 2) Mg is the limiting reactant because there isn't enough of it to completely react with the amount of I₂ present. The amount of other products can be found using the amount of Mg and mole to mole ratios. This is equal to the theoretical yield: 1.92mol Mg(1mol MgI₂/1mol Mg) = 1.92mol MgI₂ Step 3) The actual yield can be divided by the theoretical yield to get the percent yield: 1.76mol MgI₂/1.92mol MgI₂ (100) = 91.66%

What volume in milliliters of 0.0120 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl?

93.8mL; The balanced reaction is 2 HCl + Ca(OH)₂ → CaCl₂ + 2H₂O Step 1)Begin by converting the milliliters of HCl to liters: 75.0mL HCl(1L/1000mL)= 0.0750L HCl Step 2)Stoichiometry can be used to determine the volume of Ca(OH)₂ needed. The balanced reaction provides the mole ratio conversion factor: 0.0750L HCl(0.0300mol HCl/1L)(1mol Ca(OH)₂/2mol HCl)(1L/0.0120mol Ca(OH)₂)=0.09375L Ca(OH)₂ Step 3)Next, convert the liters of Ca(OH)₂ to milliliters: 0.09375L Ca(OH)₂(1000mL/1L)=93.75mL Ca(OH)₂

In the redox reaction 6 Fe²⁺ + Cr₂O₇²⁻ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O, what is the reducing agent? A) Fe²⁺ B) Cr₂O₇²⁻ C) H⁺ D) Cr³⁺ E) Fe³⁺

A) Fe²⁺; Fe²⁺ increases in oxidation number from +2 to +3 and the Cr in Cr₂O₇²⁻ decreases in oxidation number from +6 to +3. Therefore Cr is being reduced, and Fe²⁺ is the reducing agent.

If you have 5 mol H₂ and 2 mol N₂, what is the limiting reagent in the reaction below? 3 H₂(g) + N₂ (g) → 2 NH₃ (g) A) H₂ B) N₂ C) NH₃ D) Both reactants are limiting.

A) H₂; The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur. 5 mol H₂ (1 mol N₂/3 mol H₂) = 1.67 mol N₂ 2 mol N₂ (3 mol H₂/1 mol N₂) = 6 mol H₂; H₂ is the limiting reactant because there isn't enough of it to completely react with the amount of N₂.

Which one of the following would form a precipitate when mixed with aqueous Ba(NO₃)₂? A) Na₂SO₄ B) KCl C) LiC₂H₃O₂ D) CsOH E) None of these would precipitate

A) Na₂SO₄; When sodium sulfate, Na₂SO₄, is mixed with barium nitrate, Ba(NO₃)₂, in aqueous solution the precipitate barium sulfate, BaSO₄, is formed. Although sulfates are typically soluble in water, BaSO₄ is an exception and therefore insoluble in water. Ba(NO₃)₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaNO₃(aq)

What is the difference between actual yield and theoretical yield? A) Actual yield has to do only with the reactants of a reaction, and theoretical yield has to do only with the products of a reaction. B) Actual yield is how much is actually produced in a reaction (it has to be given or measured), and theoretical yield is a calculation that has to be done. C) Actual yield is how much the reaction produces, and theoretical yield is how much a reaction consumes. D) Actual yield has to do with how much you can actually get out of a reaction, and theoretical yield is how much you can get out of a reaction if you have maximum starting materials.

B) Actual yield is how much is actually produced in a reaction (it has to be given or measured), and theoretical yield is a calculation that has to be done.

In the following chemical reaction, which element is the reducing agent? 2 IO₃⁻(aq) + 12 H⁺(aq) + 10 Ag(s) + 10 Cl⁻(aq) → 10 AgCl(s) + I₂(s) + 6H₂O(l) A) I B) Ag C) Cl D) H E) O

B) Ag; Reducing agents lose electrons and are oxidized. In this reaction, Ag loses an electron to become Ag⁺.

Which of the following reactions is a double displacement reaction? A) HCl(g) → H₂(g) + Cl₂(g) B) HCl(aq) + NaOH(aq) → H₂O(l) + NaCl(aq) C) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) D) HCl(g) + C₅H₁₀(g) → C₅H₁₁Cl(g)

B) HCl(aq) + NaOH(aq) → H₂O(l) + NaCl(aq); A double displacement reaction involves two ionic compounds that react and form two new ionic compounds. The cations from the reactants "switch" places. The correct answer is HCl(aq) + NaOH(aq) → H₂O(l) + NaCl(aq).

Which of the following reactions is an acid-base reaction? A) AgNO₃(aq) + KF(aq) → AgF(s) + KNO₃(aq) B) NH₃(aq) + HF(aq) → NH₄⁺(aq) + F⁻(aq) C) NO₂⁻(aq) + Al(s) → NH₃(g) + AlO₂⁻(aq) D) CH₄(g) + 2O₂(g) → 2H₂O(g) + CO₂(g)

B) NH₃(aq) + HF(aq) → NH₄⁺(aq) + F⁻(aq); NH₃(aq) + HF(aq) → NH₄⁺(aq) + F⁻(aq) is an acid-base reaction. It involves the transfer of a proton from one species (the acid, in this case HF) to another (the base, in this case NH₃).

Which choice incorrectly identifies the oxidation number (O.N.) for the given species? A) Co³⁺, O.N. = +3 B) Br-, O.N. = -1 C) H₂, O.N. = +1 D) As³⁻, O.N. = -3 E) Si, O.N. = 0

C) H₂, O.N. = +1; The oxidation number of the elemental form of an element is 0, so O.N. = +1 is incorrect for H₂.

In the following chemical reaction, which element is the oxidizing agent? CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) A) C B) H C) O D) There is no oxidizing agent in this reaction.

C) O; Oxidizing agents gain electrons and are reduced. In this reaction, O gains two electrons to become O²⁻.

What type of reaction is represented by the following equation: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). A) combination B) decomposition C) double displacement D) single displacement

C) double displacement; This is a double displacement reaction, which involves two ionic compounds that react and form two new ionic compounds. The cations from the reactants "switch" places.

What type of reaction is represented by the following equation: CoCl₂(aq) + 2NaOH(aq) → Co(OH)₂(s) + 2NaCl(aq) A) acid-base B) decomposition C) precipitation D) redox (oxidation-reduction)

C) precipitation; This is a precipitation reaction,which involves the formation of an insoluble solid and an aqueous solution.

Combustion of an unknown compound containing only carbon and hydrogen produces 54.9 g of CO₂ and 45.1 g of H₂O. What is the empirical formula of the compound? A) C₆H₅ B) CH₂ C) CH D) CH₄

D) CH₄; Step 1)calculate number of moles of carbon dioxide and water from the given masses: 54.9g CO₂(1mol CO₂/44.01g CO₂)=1.25mol CO₂ 45.1g H₂O(1mol H₂O/18.02g H₂O=2.50mol H₂O Step 2)For every 1 mole of carbon dioxide, there are two moles of water. The combustion equation will look like this:CxHy+zO₂→CO₂+2H₂O Since there is 1 mole of C, 4 moles of H and 4 moles of O on the product side, there must be the same amounts on the reactant side: CH₄+2O₂→CO₂+2H₂O; Because CH₄ cannot be reduced any further, this is also its empirical formula.

When propanol (C₃H₇OH) is burned in excess oxygen, what is/are the product(s) of this reaction? A) CO₂ B) H₂O, O₂ C) H₂O D) CO₂, H₂O

D) CO₂, H₂O; The products of a combustion reaction are carbon dioxide and water.

Which choice correctly identifies the oxidation numbers (O.N.) for each element in Ca(NO₃)₂? A) Ca = 0, N = 0, O = 0 B) Ca = 0, N = +5, O = -2 C) Ca = +2, N = +5, O = -6 D) Ca = +2, N = +5, O = -2 E) Ca = +4, N = +5, O = -2

D) Ca = +2, N = +5, O = -2; The charge on the NO₃⁻ ion is -1, so Ca must have a charge of +2 and an O.N. of +2. O always has a O.N. of -2 for compounds other than peroxides, which means N must have an O.N. of +5.

If 8.00 moles of NH₃ of and 10.00 moles of O₂ react in the following reaction, how many moles of which reactant will be left over? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) A) 2.00 mol O₂ B) 2.00 mol NH₃ C) 8.00 mol NO D) no moles will be left over of either reactant because both are limiting.

D) no moles will be left over of either reactant because both are limiting; The limiting reactant can be determined by calculating the amount of the other reactant that would be necessary for the reaction to occur: 8.00mol NH₃ (5mol O₂/4mol NH₃)= 10.00mol O₂ 10.00mol O₂ (4 mol NH₃/5mol O₂)= 8.00mol NH₃ No reactant will be leftover because both are limiting and fully used.

What type of reaction is represented by the following equation: C₄H₈(g) + 6O₂(g) → 4CO₂(g) + 4H₂O(g) A) acid-base B) decomposition C) precipitation D) redox (oxidation-reduction)

D) redox (oxidation-reduction); This is a redox equation, which involves the oxidation (loss of electrons) and reduction (gain of electrons) between species in the reaction.

What type of reaction is represented by the following equation: SiCl₄(l) + 2Mg(s) → 2MgCl₂(s) + Si(s) A) combination B) decomposition C) double displacement D) single displacement

D) single displacement; This is a single displacement reaction,which involves one element taking the place of another in a reaction. A metal cation is replaced by another metal in the ionic compound and the cation becomes a neutral metal.

What are the oxidation numbers of each element in the arsenate ion, AsO₄³⁻? A) As = 0, O = -3 B) As = 0, O = -2 C) As = -3, O = -2 D) As = +3, O = -2 E) As = +5, O = -2

E) As = +5, O = -2; O always has an oxidation number of -2 for compounds other than peroxides, so As must have an oxidation number of +5 to give a -3 charge on the ion.

Balance the following chemical equation (if necessary): NH₄NO₂(s)→ N₂(g) + H₂O(l)

NH₄NO₂(s)→ N₂(g) + 2 H₂O(l)

Write the balanced molecular chemical equation for the reaction in aqueous solution for ammonium sulfate and iron(II) chloride. If no reaction occurs, write NR.

NR

Write a balanced chemical equation based on the following description: gaseous nitrogen and hydrogen react to form gaseous ammonia

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Complete and balance the following half-reaction in basic solution: N₂(g) → NH₃(g)

N₂(g) + 6 H₂O(l) + 6 e⁻ → 2 NH₃(g) + 6 OH⁻(aq)

Complete the balanced molecular chemical equation for the reaction below. If no reaction occurs, write NR after the reaction arrow. Pb(NO₃)₂(aq) + KI(aq) →

Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq)

Balance the following chemical equation (if necessary): SO₄²⁻(aq) + Ba²⁺(aq) ⇌ BaSO₄(s)

SO₄²⁻(aq) + Ba²⁺(aq) ⇌ BaSO₄(s)

A 10.68 g sample of compound that contains only carbon, hydrogen, and oxygen was analyzed through combustion analysis and yielded 16.01 g of CO₂ and 4.37 g of H₂O. What is this compound's empirical formula?

The conservation of mass can be used to find the mass of oxygen that was not a result of the O₂ reactant. Then the amount of oxygen in each of the products can be calculated in the same way as carbon and hydrogen. 6)16.01+4.37-10.68=9.7gO(1mol O/16.00g O)=0.60625mol O reactant 7)44.01gCO₂/32.00gO=16.01gCO₂/x 8)(32.00gOx16.01gCO₂)/44.01gCO₂=11.64gO=(1molO/16.00gO)=0.7276molO 9)18.02gH₂O/16.00gO=4.37gH₂O/x 10)(16.00gOx4.37gH₂O)/18.02gH₂O= 3.88gO(1molO/16.00gO)=0.243molO The number of moles of oxygen can be determined using the law of conservation of mass and the ratio between oxygen and carbon can be determined. 11)0.7276+0.243-0.60625=0.364molO in compound 0.364molO/0.364molC=1molO/1molC Therefore, if there are 4H for every 3C, and 1O for every 1C, the empirical formula would be C₃H₄O₃.