Chemical Reactions, Energy Changes, and Redox Reactions: Questions and Examples

A 4.33 g sample of an unknown alkali hydroxide compound is dissolved completely in water. A sufficient solution of copper (II) nitrate is added to the hydroxide solution such that it will fully precipitate copper (II) hydroxide via the following reaction: Cu²⁺ + 2OH⁻ → Cu(OH)₂(s) After the precipitate is filtered and dried, its mass is found to be 3.81 g. Is the original alkali hydroxide sample most likely LiOH, NaOH, or KOH?

(3.81 g Cu(OH)₂) x (1 mol/97.55 g) = 0.0391 mol Cu(OH)₂ (0.0391 mol Cu(OH)₂) x (2 mol OH⁻/1 mol Cu(OH)₂) x (17.02 g OH⁻/1 mol) = 1.33 g OH⁻ 1.33 g OH- in precipitate = 1.33 g OH⁻ in original sample 1.33 g OH⁻/4.33 g sample = 30.7% *Find mass percent of hydroxide in all possible compounds: LiOH: 17.02 g/23.96 g = 71.0% NaOH: 17.02 g/40.02 = 42.5% KOH: 17.02 g/56.12 g = 30.3% Since the mass percent of hydroxide in the sample is closest to the mass percent of hydroxide in KOH, KOH is likely the identity of the unknown compound.

How many minutes will take to plate out 40.00 g of Ni form a solution of NiSO₄ using a current of 3.450 amp?

(40.00 g Ni)(58.69 g/1 mol) = 0.6815 moles Ni Ni²⁺ + 2e⁻ → Ni (o.6815 mol Ni)(2 mol e⁻/1 mol Ni) = 1.363 moles e⁻ (1.363 mol e⁻)(96,500 C/1 mol) = 131500 coulombs 1 ampere = 1 C/sec 3.450 amp = 131500 coulombs/ x seconds 131500/3.450 = 38120 seconds (38120 seconds)(1 minutes/60 seconds) = 635.3 minutes

Estimate the energy change for the following reaction using bond energies: N₂(g) + 3H₂(g) →2NH₃(g) List of bond energies: H-H single bond = 436 kj/mol N-N double bond = 418 kj/mol N-N triple bond = 941 kj/mol N-H single bond = 391 kj/mol H-F single bond = 567 kj/mol

*First draw Lewis structures: -N₂ has one N-N triple bond -H₂ has one H-H single bond -2NH₃ has three N-H single bonds ∆H = H_broken - H_formed ∆H = (941 + (3 x 436)) - (2 x 3 x 391) ∆H = 2249 - 2346 ∆H = -97 kj/mol

Balance the following chemical equation: Al + CuSO₄ → Al₂(SO₄)₃ + Cu

2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

Balance the following chemical equation: C₄H₁₀ + O₂ → CO₂ + H₂O

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Balance the following chemical equation: Fe₂O₃ + C → Fe₃O₄ + CO

3Fe₂O₃ + C → 2Fe₃O₄ + CO Note: Balance Fe based on subscripts instead of O because O is also in CO

Balance the following chemical equation: Na₂SO₃ + S₈ → Na₂S₂O₃

8Na₂SO₃ + S₈ → 8Na₂S₂O₃

Balance the following redox reaction: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺

Half-reactions: Reduction: 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O Oxidation: Fe²⁺ → Fe³⁺ + e⁻ *Use least common multiple to get same number of electrons = 6 Reduction: 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O Oxidation: 6Fe²⁺ → 6Fe³⁺ + 6e⁻ *Add equations together and cancel anything on both sides Balanced Equation: 6Fe²⁺ +14H⁺ +Cr₂O₇⁻ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound?

Molar Mass CO₂ = 44.01 grams/mole Molar Mass C = 12.01 grams/mole Carbon: (0.3664 g) x (12.01/44.01) = 0.1000 grams C Molar Mass H₂O = 18.02 grams/mole Molar Mass H = 1.008 x 2 = 2.016 grams/mole Hydrogen: (0.1500 g) x (2.016/18.02) = 0.01678 grams H 0.01678 + 0.1000 = 0.1168 0.2500 - 0.1168 = 0.1332 grams O Carbon: (0.1000 g)(1 mol/12.01 g) = 0.008326 moles Hydrogen: (0.01678 g)(1 mol/1.008 g) = 0.01665 moles Oxygen: (0.1332 g)(1 mol/16.00 g) = 0.008325 moles Smallest = 0.008325 Carbon: 0.008326 / 0.008325 ~ 1 Hydrogen: 0.01665 / 0.008325 = 2 Oxygen: 0.008325 / 0.008325 = 1 Empirical formula: CH₂O

Determine the oxidation number of each element in (NH₄)₂SO₄

Nitrogen is a non-metal so H = +1 O = -2 NH₄ has a charge of +1 (4 x +1) = +4 +1 - +4 = N = -3 SO₄ has a charge of -2 (4 x -2) = -8 -2 - (-8) = -2 + 8 = S = +6

NH₃ + O₂ → N₂ + H₂O If the equation above were balanced with the lowest whole number coefficients, what would the coefficient of NH₃ be: A) 1 B) 2 C) 3 D) 4

Note: Since options are given, each one can be tried to see which works. Answer: D) 4 *The balanced equation would be 4NH₃ + 3O₂ → 2N₂ + 6H₂O

A 50.0 mL solution of sodium oxalate, Na₂C₂O₄ is poured into an Erlenmeyer flask. An acidified solution of 0.135 M potassium permanganate is titrated into the flask while the solution is swirled on a stir station and heated. The solution in the flask is originally colorless but turns pink after 14.56 mL of potassium permanganate is added. Given the following half-reactions, what is the concentration of the oxalate solution? Reduction: 8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O(l) Oxidation: C₂O₄²⁻ → 2CO₂(g) + 2e⁻

Overall reaction: 16H⁺ + 5C₂O₄²⁻ + 2MnO₄⁻ → 2Mn²⁺ + 8H₂O(l) + 5CO₂(g) (0.135 M MnO₄⁻) / (0.01456 L MnO₄⁻) = 0.00197 moles MnO₄⁻ (0.00197 moles) x (5 mol C₂O₄²⁻/2 mol MnO₄⁻) = 0.00493 moles C₂O₄²⁻ 0.00493 moles/0.0500 L = [C₂O₄²⁻] = [Na₂C₂O₄] = 0.0986 M

Identify the substance being oxidized and the substance being reduced: 2As(s) + 3 Cl₂(g) → 2AsCl₃

Oxidized: As Reduced: Cl₂

If 24 grams of sodium chloride reacts with an excess amount of magnesium oxide, how many grams of sodium oxide will be produced?

Reaction: 2NaCl + MgO → Na₂O + MgCl₂ Molar Mass of NaCl = 58.4 g/mol (24 g NaCl) x (1 mol NaCl/58.4 g) x (1 mol Na₂O/2 mol NaCl) x (61.98 g/1 mol Na₂O) = 12.7 g Na₂O

Would NaBr be soluble in water?

Yes, according to solubility rules, compounds with an alkali metal cation are always soluble. For a more detailed explanation, when NaBr dissolves in water, the positive Na⁺ cations are attracted to the negative oxygen ends of the water molecules. The negative Br⁻ anions are attracted to the positive hydrogen ends of the water molecules. Ionic substances like NaBr dissolve in water when these attractions of ions to dipoles of water molecules exceed attraction of ions to each other.