Aluminum reacts with chlorine to form aluminum chloride.

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 reacts with chlorine to form aluminum chloride

Balanced equation:
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’ll use stoichiometry to find the theoretical yield of AlCl₃ and then use the formula:

% Yield = (Actual Yield / Theoretical Yield) × 100

1. 20.00 g of Cl₂ produces 24.75 g of AlCl₃

• Moles of Cl₂ = 20.00 g ÷ 70.90 g/mol = 0.282 mol
• From the equation, 3 mol Cl₂ produces 2 mol AlCl₃
• So, 0.282 mol Cl₂ produces:
  (2/3)×0.282=0.188molAlCl3(2/3) × 0.282 = 0.188 mol AlCl₃(2/3)×0.282=0.188molAlCl3​
• Mass of AlCl₃ = 0.188 mol × 133.33 g/mol = 25.06 g
• % Yield = (24.75 ÷ 25.06) × 100 = 98.71%

2. 33.50 g Al produces 164.5 g AlCl₃

• Moles of Al = 33.50 ÷ 26.98 = 1.242 mol
• From the equation, 2 mol Al produces 2 mol AlCl₃ (1:1)
• So, 1.242 mol Al → 1.242 mol AlCl₃
• Mass = 1.242 × 133.33 = 165.6 g
• % Yield = (164.5 ÷ 165.6) × 100 = 99.36%

3. 145.0 g Al produces 679.4 g AlCl₃

• Moles of Al = 145.0 ÷ 26.98 = 5.374 mol
• So, AlCl₃ = 5.374 mol
• Mass = 5.374 × 133.33 = 716.8 g
• % Yield = (679.4 ÷ 716.8) × 100 = 94.81%

4. 5.000 g Cl₂ produces 6.000 g AlCl₃

• Moles Cl₂ = 5.000 ÷ 70.90 = 0.0705 mol
• Moles AlCl₃ = (2/3) × 0.0705 = 0.047 mol
• Mass = 0.047 × 133.33 = 6.27 g
• % Yield = (6.000 ÷ 6.27) × 100 = 95.72%

Matching:

• 20.00 g Cl₂ → 98.71%
• 33.50 g Al → 99.36%
• 145.0 g Al → 94.81%
• 5.000 g Cl₂ → 95.72%

Part B: Hydrogen and nitrogen form ammonia

Balanced equation:
3 H₂ + N₂ → 2 NH₃

Use stoichiometric ratios:

• 3 mol H₂ per 2 mol NH₃
• 1 mol N₂ per 2 mol NH₃

1. Amount of H₂ to react with 8.2 mol N₂

• 1 N₂ : 3 H₂ → 3 × 8.2 = 24.6 mol → Option 4

2. Amount of N₂ to produce 50.4 mol NH₃

• 2 NH₃ : 1 N₂ → 50.4 ÷ 2 = 25.2 mol → Option 3

3. NH₃ produced from 7.0 mol N₂

• 1 N₂ → 2 NH₃ → 2 × 7.0 = 14.0 mol → Option 6

4. H₂ to produce 44.0 mol NH₃

• 2 NH₃ : 3 H₂ → (44.0 × 3) ÷ 2 = 66.0 mol → Option 2

5. NH₃ produced from 12.7 mol H₂

• 3 H₂ → 2 NH₃ → (12.7 × 2) ÷ 3 = 8.47 mol → Option 1

6. N₂ to react with 5.0 mol H₂

• 3 H₂ : 1 N₂ → 5.0 ÷ 3 = 1.67 mol → Option 5

Final Matching:

• H₂ with 8.2 mol N₂ → 24.6 mol → Option 4
• N₂ to make 50.4 mol NH₃ → 25.2 mol → Option 3
• NH₃ from 7.0 mol N₂ → 14.0 mol → Option 6
• H₂ for 44.0 mol NH₃ → 66.0 mol → Option 2
• NH₃ from 12.7 mol H₂ → 8.47 mol → Option 1
• N₂ for 5.0 mol H₂ → 1.67 mol → Option 5