Question

Type Calculations. Given the balanced equation: 2 Al + 3 H2SO4---> Al2(SO4)3 + 3 H2 Molar mass (g/mol): Al=26.98; H2SO4=98.08; Al2(SO4)3= 342.15; H2=2.02 A) How many mole of H2 gas can be produced when 33.8 g of Al are consumed? B) When 1.60 mol of H2SO4 are used in a reaction, how many grams of Al2(SO4)3 can be produced? C) For part B,if actual yield for Al2(SO4)3 is 100.0 g, what is percent yield?

Answer 1

A. 1.88 mol H₂

B. 182 g Al₂(SO₄)₃

C. 54.8%

Step-by-step explanation:

2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂

A. Convert grams of Al to moles. The molar mass is 26.98 g/mol.

(33.8 g)/(26.98 g/mol) = 1.253 mol Al

Use stoichiometry to convert moles of Al to moles of H₂. Looking at the equation, you can see that for every 2 mol of Al consumed, 3 moles of H₂ is produced. Use this relationship.

(1.253 mol Al) × (3 mol H₂)/(2 mol Al) = 1.879 mol H₂

You will produce 1.88 mol of H₂ gas.

B. Again, use stoichiometry. For every 3 moles of H₂SO₄ consumed, 1 mole of Al₂(SO₄)₃ is produced.

(1.60 mol H₂SO₄) × (1 mol Al₂(SO₄)₃/3 mol H₂SO₄) = 0.533 mol Al₂(SO₄)₃

Convert moles of Al₂(SO₄)₃ to grams. The molar mass is 342.15 g/mol.

(0.533 mol) × (342.15 g/mol) = 182.48 g Al₂(SO₄)₃

You will produce 182 g of Al₂(SO₄)₃.

C. Calculate percent yield by dividing the actual yield by the theoretical yield. Multiply by 100%.

(100.0/182.48) × 100% = 54.8%

The percent yield is 54.8%.