Answer to Question #325555 in Chemistry for Liky

Question

If you react 5.625 L of H2S in excess oxygen, how many grams of water will be produced

Answer 1

Solution:

At STP, one mole of any gas occupies a volume of 22.4 L.

Therefore,

Moles of H₂S = (5.625 L H₂S) × (1 mol H₂S / 22.4 L H₂S) = 0.2511 mol H₂S

Balanced chemical equation:

2H₂S + 3O₂ → 2SO₂ + 2H₂O

According to stoichiometry:

2 mol of H₂S produce 2 mol of H₂O

Thus, 0.2511 mol of H₂S produce:

(0.2511 mol H₂S) × (2 mol H₂O / 2 mol H₂S) = 0.2511 mol H₂O

The molar mass of water (H₂O) is 18.0153 g/mol

Therefore,

Mass of H₂O = (0.2511 mol H₂O) × (18.0153 g H₂O / 1 mol H₂O) = 4.524 g H₂O

Mass of H₂O = 4.524 g

Answer: 4.524 grams of water (H₂O) will be produced