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) A) 1.48 × 10²⁴molecules H₂S B) 9.89 × 10²³molecules H₂S C) 5.06 × 10²⁵molecules H₂S D) 3.17 × 10²⁵molecules H₂S E) 2.44 × 10²³molecules H₂S

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First, we need to find the moles of sulfur using its molar mass: $$79.0 ext{ g S} imes \frac{1 ext{ mol S}}{32.066 ext{ g S}} = 2.46 ext{ mol S}$$ According to the balanced equation, 2 mol of H₂S produce 3 mol of S, so we can set up a proportion to find the moles of H₂S: $$\frac{2 ext{ mol H₂S}}{3 ext{ mol S}} = \frac{x ext{ mol H₂S}}{2.46 ext{ mol S}}$$ Solving for x, we get: $$x = \frac{2.46 ext{ mol S} imes 2 ext{ mol H₂S}}{3 ext{ mol S}} = 1.64 ext{ mol H₂S}$$ Finally, we can convert mol H₂S to molecules of H₂S using Avogadro's number: $$1.64 ext{ mol H₂S} imes 6.022 imes 10^{23} ext{ molecules/mol} = 9.89 imes 10^{23} ext{ molecules of H₂S}$$ Therefore, the answer is choice B.

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