Question

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂.

2 H₂S(g) + SO₂(g) → 3 S(s) + 2 H₂O(l)

Answer 1

To produce 79.0 g of sulfur, approximately 9.89 x 10²³ molecules of H₂S are required, using the balanced chemical reaction and stoichiometric calculations.

Step-by-step explanation:

To determine how many molecules of H₂S are required to form 79.0 g of sulfur, we first need to use the provided chemical reaction, which is already balanced:

2 H₂S(g) + SO₂(g) → 3 S(s) + 2 H₂O(l)

According to the reaction, 2 moles of H₂S produce 3 moles of sulfur (S). To find out how many moles of H₂S are needed to form 79.0 g of sulfur, we first need to calculate the moles of sulfur produced.

Step 1: Convert grams of sulfur to moles.

1. Molar mass of sulfur (S) = 32.07 g/mol
2. Moles of sulfur = mass of sulfur / molar mass of sulfur = 79.0 g / 32.07 g/mol = 2.464 mol

Step 2: Use the stoichiometry of the reaction to find moles of H₂S.

1. Moles of H₂S required = (2 moles H₂S / 3 moles S) × 2.464 mol = 1.6427 mol

Step 3: Convert moles of H₂S to molecules.

1. Number of molecules in 1 mole = Avogadro's number = 6.022 × 10²³ molecules/mol
2. Molecules of H₂S required = 1.6427 mol × 6.022 × 10²³ molecules/mol = 9.89 × 10²³ molecules

Therefore, approximately 9.89 × 10²³ molecules of H₂S are needed to produce 79.0 g of sulfur.