Answer to Question #290494 in Chemistry for thaiana

Question

When 9.0 g of hydrogen gas, H₂, reacts with oxygen gas, O₂, 73.0 g of water is produced.    2H2 + O2 --> 2H2O a) What is the theoretical yield of water? b) What is the percentage yield?

Answer 1

Solution:

Calculate the moles of hydrogen gas (H₂):

The molar mass of H₂ is 2.0 g/mol

Hence,

Moles of H₂ = (9.0 g H₂) × (1 mol H₂ / 2.0 g H₂) = 4.5 mol H₂

Balanced chemical equation:

2H₂ + O₂ → 2H₂O

According to stoichiometry:

2 mol of H₂ produces 2 mol of H₂O

Thus, 4.5 mol of H₂ produces:

(4.5 mol H₂) × (2 mol H₂O / 2 mol H₂) = 4.5 mol H₂O

Calculate the theoretical yield of water (H₂O):

The molar mass of H₂O is 18.0 g/mol

Hence,

Mass of H₂O = (4.5 mol H₂O) × (18.0 g H₂O / 1 mol H₂O) = 81.0 g H₂O

The theoretical yield of water (H₂O) is 81.0 grams

Calculate the percentage yield:

%yield = (actual yield / theoretical yield) × 100%

%yield = (73.0 g / 81.0 g) × 100% = 90.1%

The percentage yield is 90.12%

Answer:

The theoretical yield of water is 81.0 grams

The percentage yield is 90.12%