Aluminum reacts with chlorine to form aluminum chloride. The equation for this reaction is 2 Al(s) + 3 Cl2(g) -> 2 AlCl3(s). Match each situation to the corresponding percentage yield. Answer choices may be used only once. Use the molar mass values given here. Al = 26.98 g/mol Cl2 = 70.90 g/mol AlCl3 = 133.33 g/mol – 20.00 g of chlorine was reacted with excess aluminum to produce 24.75 g of aluminum chloride. – 33.50 g of aluminum was reacted with excess chlorine to produce 164.5 g of aluminum chloride. – 145.0 g of aluminum was reacted with excess chlorine to produce 679.4 g of aluminum chloride. – 5.000 g of chlorine was reacted with excess aluminum to produce 6.000 g of aluminum chloride. Options: 1. 95.72% 2. 94.81% 3. 98.71% 4. 99.36% b) Hydrogen and nitrogen react to form ammonia. The chemical equation for this reaction is shown here: 3 H2(g) + N2(g) -> 2 NH3(g). Match each situation to the corresponding amount. Answer choices may be used only once. – Amount of H2 that will react with 8.2 mol of N2. – Amount of N2 that will produce 50.4 mol of NH3. – Amount of NH3 produced when 7.0 mol of N2 is consumed. – Amount of H2 that will produce 44.0 mol of NH3. – Amount of NH3 produced when 12.7 mol of H2 is consumed. – Amount of N2 that will react with 5.0 mol of H2. Options: 1. 8.47 mol 2. 66.0 mol 3. 25.2 mol 4. 24.6 mol 5. 1.67 mol 6. 14.0 mol
The Correct Answer and Explanation is:
Part A: Aluminum and Chlorine Reaction
Reaction:
2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s)
Molar masses:
Al = 26.98 g/mol
Cl₂ = 70.90 g/mol
AlCl₃ = 133.33 g/mol
We use stoichiometry to determine theoretical yields and then compare with actual yields to calculate percent yield using the formula:
Percent yield = (Actual yield / Theoretical yield) × 100
1. 20.00 g of chlorine reacted with excess aluminum to produce 24.75 g of aluminum chloride.
Moles of Cl₂ = 20.00 g ÷ 70.90 g/mol = 0.2821 mol
From stoichiometry: 3 mol Cl₂ produce 2 mol AlCl₃
So 0.2821 mol Cl₂ produces:
(2/3) × 0.2821 = 0.1881 mol AlCl₃
Mass = 0.1881 mol × 133.33 g/mol = 25.07 g (theoretical yield)
% yield = (24.75 ÷ 25.07) × 100 = 98.71%
→ Match: 3. 98.71%
2. 33.50 g of aluminum reacted with excess chlorine to produce 164.5 g of aluminum chloride.
Moles of Al = 33.50 ÷ 26.98 = 1.2417 mol
From stoichiometry: 2 mol Al produce 2 mol AlCl₃ → 1:1
So, 1.2417 mol Al → 1.2417 mol AlCl₃
Mass = 1.2417 × 133.33 = 165.55 g (theoretical yield)
% yield = (164.5 ÷ 165.55) × 100 = 99.36%
→ Match: 4. 99.36%
3. 145.0 g of aluminum reacted with excess chlorine to produce 679.4 g of aluminum chloride.
Moles of Al = 145.0 ÷ 26.98 = 5.3733 mol
So 5.3733 mol Al → 5.3733 mol AlCl₃
Mass = 5.3733 × 133.33 = 716.37 g
% yield = (679.4 ÷ 716.37) × 100 = 94.81%
→ Match: 2. 94.81%
4. 5.000 g of chlorine reacted with excess aluminum to produce 6.000 g of aluminum chloride.
Moles Cl₂ = 5.000 ÷ 70.90 = 0.0705 mol
Moles AlCl₃ = (2/3) × 0.0705 = 0.047 mol
Mass = 0.047 × 133.33 = 6.266 g
% yield = (6.000 ÷ 6.266) × 100 = 95.72%
→ Match: 1. 95.72%
Part B: Hydrogen and Nitrogen to Ammonia
Reaction:
3 H₂ + N₂ → 2 NH₃
Use mole ratios from the equation to match:
1. Amount of H₂ that will react with 8.2 mol of N₂
3 mol H₂ per 1 mol N₂ → 3 × 8.2 = 24.6 mol
→ Match: 4. 24.6 mol
2. Amount of N₂ that will produce 50.4 mol of NH₃
From 1 mol N₂ → 2 mol NH₃
N₂ = 50.4 ÷ 2 = 25.2 mol
→ Match: 3. 25.2 mol
3. Amount of NH₃ produced when 7.0 mol of N₂ is consumed
2 mol NH₃ per 1 mol N₂ → 2 × 7.0 = 14.0 mol
→ Match: 6. 14.0 mol
4. Amount of H₂ that will produce 44.0 mol of NH₃
From 2 mol NH₃ needs 3 mol H₂
H₂ = 44.0 × (3/2) = 66.0 mol
→ Match: 2. 66.0 mol
5. Amount of NH₃ produced when 12.7 mol of H₂ is consumed
From 3 mol H₂ → 2 mol NH₃
NH₃ = 12.7 × (2/3) = 8.47 mol
→ Match: 1. 8.47 mol
6. Amount of N₂ that will react with 5.0 mol of H₂
From 3 mol H₂ → 1 mol N₂
N₂ = 5.0 ÷ 3 = 1.67 mol
→ Match: 5. 1.67 mol
