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) + 2H2O(l)
Answer
9.89 × 1023 molecules H2S
5.06 × 1025 molecules H2S
2.44 × 1023 molecules H2S
1.48 × 1024 molecules H2S
3.17 × 1025 molecules H2S

Answer 1

2 H2S(g) + SO2(g) ? 3 S(s) + 2H2O(l)

Mol-mol relationsship between H2S & S:
2 mol H2S / 3 mol S

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Calculation:
Excess SO2 ==> limited H2S produces S

79.0 g S
x (1 mol S / 32.065 g S) <--- MW of S
x (2 mol H2S / 3 mol S) <--- mol-mol relation from equation
x (34.081 g H2S / 1 mol H2S) <--- MW of H2S
(notice all units cancel except g H2S)
= 55.9779 g H2S

*** Answer: 56.0 g H2S ***

molecular weight of H2S is 34.08

so answer is (56/34.08)*6.023 * 10^-23

 = 9.89 × 1023 molecules H2S