How many molecules of H2S are required to form 79.0 g of sulfur according to the reaction: 2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(l)? Assume excess SO2.

Use the stoichiometry of the balanced chemical equation: \( 2 \text{ H}_2\text{S} \rightarrow 3 \text{ S} \). This means 2 moles of \( \text{H}_2\text{S} \) produce 3 moles of \( \text{S} \). Calculate the moles of \( \text{H}_2\text{S} \) required using the ratio: \( \text{moles of H}_2\text{S} = \frac{2}{3} \times \text{moles of S} \).

Convert the moles of \( \text{H}_2\text{S} \) to molecules using Avogadro's number, \( 6.022 \times 10^{23} \text{ molecules/mol} \). Use the formula: \( \text{molecules of H}_2\text{S} = \text{moles of H}_2\text{S} \times 6.022 \times 10^{23} \).