Chp. 8 Chemical Equations Calculations

When exposed to air, aluminum metal, Al, reacts with oxygen, O2, to produce a protective coating of aluminum oxide, Al2O3, which prevents the aluminum from rusting underneath. The balanced reaction is shown here: 4Al+3O2→2Al2O3 A) What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen? B) In Part A, we saw that the theoretical yield of aluminum oxide is 0.700 mol . Calculate the percent yield if the actual yield of aluminum oxide is 0.441 mol .

A) 0.700 mol B) 63.0%

A) Balance the chemical reaction equation P4(s)+Cl2(g)→PCl5(g) B) How many moles of PCl5 can be produced from 25.0 g of P4 (and excess Cl2)? C) How many moles of PCl5 can be produced from 55.0 g of Cl2 (and excess P4)? D) What mass of PCl5 will be produced from the given masses of both reactants?

A) 1,10,4 B) 0.807 mole C) 0.310 mole D) 64.6 g

Consider a situation in which 161 g of P4 are exposed to 176 g of O2. A) What is the maximum amount in moles of P2O5 that can theoretically be made from 161 g of P4 and excess oxygen? B) What is the maximum amount in moles of P2O5 that can theoretically be made from 176 g of O2 and excess phosphorus? C) In Part A, you found the amount of product (2.60 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the amount of product (2.20 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus. Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus and oxygen. D) What is the percent yield if the actual yield from this reaction is 247 g ?

A) 2.60 mol B) 2.20 mol C) 2.20 mol D) 79.0%

Ammonium nitrate reacts explosively upon heating to form nitrogen gas, oxygen gas, and gaseous water. A) Write a balanced equation for this reaction. B) Determine how much oxygen in grams is produced by the complete reaction of 2.95 kg of ammonium nitrate.

A) 2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g) B) m= 590 g

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation: 2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g) The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react. Also, notice that the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce. But be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios. A) What is the molar mass of butane, C4H10? B) Calculate the mass of water produced when 9.86 g of butane reacts with excess oxygen. C) Calculate the mass of butane needed to produce 49.8 g of carbon dioxide.

A) 58.12 g/mol B) 15.3 g C) 16.4 g

Under certain circumstances, carbon dioxide, CO2(g), can be made to react with hydrogen gas, H2(g), to produce methane, CH4(g), and water vapor, H2O(g): CO2(g)+4H2(g)→CH4(g)+2H2O(g) A) How many moles of methane are produced when 59.6 moles of carbon dioxide gas react with excess hydrogen gas? B) How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 99.1 moles of water vapor?

A) 59.6 mol B) 198 mol

In this simulation, select the Sandwiches icon where you can set the number of bread slices, cheese slices, and meat slices to prepare a sandwich. You can then move to the Molecules version where you will be able to relate the analogy to one of three specific chemical reactions. A) In the Sandwiches mode, select the "Cheese sandwich" option and observe the equation given for the preparation of a cheese sandwich. When the number of bread slices and cheese slices is set to "0," you will see that there is "no reaction." Now, in the equation, set the number of bread slices to "2" and the number of cheese slices to "1." You will see that the product formed by these three elements is one cheese sandwich with two slices of bread and one slice of cheese. This equation can be written as 2 bread + cheese → cheese sandwich Classify the following combination of bread and cheese according to the limiting ingredient in the preparation of the cheese sandwich mentioned above. B) In the Sandwiches, mode select the "Cheese sandwich" option and observe the equation given for the preparation of a cheese sandwich. In the equation, set the number of bread slices to "2" and the number of cheese slices to "1." You will see that the product formed by these three ingredients is one cheese sandwich. Suppose you have 44 bread slices and 34 cheese slices. How many cheese sandwiches can you make? C) Go to the Molecules mode by selecting the icon at the bottom. Notice that there are three real reactions: the synthesis of water, the synthesis of ammonia, and the combustion of methane. Select each of the reactions, and observe the reactants taking part in the reactions. Suppose you are carrying out each of these reactions starting with 5 mol of each reactant. Observe that some reactants are in excess whereas some reactants limit the amount of products formed. Classify each of the reactants as a limiting reactant or an excess reactant for a reaction starting with 5 mol of each reactant. D) Find the number of moles of water that can be formed if you have 118 mol of hydrogen gas and 54 mol of oxygen gas.

A) Bread is the limiting ingredient: 4 bread slices and 5 cheese sliced; 8 bread slices and 5 cheese slices; 6 bread slices and 4 cheese slices Cheese is the limiting ingredient: 4 bread slices and 1 cheese slice; 6 bread slices and 2 cheese slices; 8 bread slices and 3 cheese slices There is no limiting ingredient: 6 bread slices and 3 cheese slices; 2 bread slices and 1 cheese slice B) 22 sandwiches C) Limiting reagent: H2 in formation of water; H2 in formation of ammonia; O2 in combustion of methane Excess reagent: N2 in formation of ammonia; O2 in formation of water; CH4 in combustion of methane D) 108 mol

Aspirin can be made in the laboratory by reacting acetic anhydride (C4H6O3) with salicylic acid (C7H6O3) to form aspirin (C9H8O4) and acetic acid (C2H4O2). The balanced equation is: C4H6O3+C7H6O3→C9H8O4+C2H4O2.In a laboratory synthesis, a student begins with 5.00 mL of acetic anhydride (density = 1.08 g / mL) and 2.08 g of salicylic acid. Once the reaction is complete, the student collects 2.50 g of aspirin. A) Determine the limiting reactant for the reaction. B) Determine the theoretical yield of aspirin for the reaction. C) Determine the percent yield for the reaction.

A) C7H6O3 B) m= 2.71 g C) percent yield= 92.3%

Urea (CH4N2O), a common fertilizer, can be synthesized by the reaction of ammonia (NH3) with carbon dioxide: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 70.2 kg of ammonia with 105 kg of carbon dioxide. A) Determine the limiting reactant. B) Determine the theoretical yield of urea. C) Determine the percent yield for the reaction.

A) NH3 B) mCH4N2O= 124 kg C) 70.7%

As we have seen, scientists have grown increasingly worried about the potential for global warming caused by increasing atmospheric carbon dioxide levels. The world burns the fossil fuel equivalent of approximately 7.50×1012 kg of petroleum per year. Assume that all of this petroleum is in the form of octane (C8H18). A) Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year. ( Hint: Begin by writing a balanced equation for the combustion of octane.) B) If the atmosphere currently contains approximately 3.50×1015 kg of CO2, how long will it take for the world's fossil fuel combustion to double the amount of atmospheric carbon dioxide?

A) mCO2= 2.31x10^13 kg B) 151 years

For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume that there is more than enough of the other reactant. C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) A) 4.4 molC3H8 B) 4.4 molC3H8 C) 0.0503 molC3H8 D) 0.0503 molC3H8 E) 4.4 molO2 F) 4.4 molO2 G) 0.0503 molO2 H) 0.0503 molO2

A) v= 13 mol CO2 B) v= 18 mol H2O C) v= 0.151 mol CO2 D) v= 0.201 mol H2O E) v= 2.6 mol CO2 F) v= 3.5 mol H2O G) v= 3.02x10^-2 mol CO2 H) v= 4.02x10^-2 mol H2O

How much hydrochloric acid in grams can be neutralized by 4.00 g of sodium bicarbonate? (Hint: Begin by writing a balanced equation for the reaction between aqueous sodium bicarbonate and aqueous hydrochloric acid.)

m= 1.74 g

The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate how many kilograms of carbon dioxide are added to the atmosphere per 7.3 kg of octane burned. ( Hint: Begin by writing a balanced equation for the combustion reaction.)

m= 23 kg

How much zinc in grams is required to make 19.7 g of hydrogen gas through this reaction?

m= 638 g

How much heat is produced by the complete combustion of 234 g of CH4? CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH degrees rxn=−802.3kJ

|Q| = 1.17×104 kJ