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Answer :
To find out how many grams of sulfur are formed when 37.4 g of water are produced, we can follow these steps using the balanced chemical equation:
1. Equation Analysis:
The balanced equation is:
[tex]\[
2 \text{H}_2\text{S(g)} + \text{SO}_2\text{(g)} \rightarrow 3 \text{S(s)} + 2 \text{H}_2\text{O(l)}
\][/tex]
2. Molar Masses:
- Molar mass of water (H₂O) ≈ 18.02 g/mol
- Molar mass of sulfur (S) ≈ 32.07 g/mol
3. Moles of Water:
First, calculate the moles of water formed:
[tex]\[
\text{moles of H}_2\text{O} = \frac{\text{mass of H}_2\text{O}}{\text{molar mass of H}_2\text{O}} = \frac{37.4 \text{ g}}{18.02 \text{ g/mol}}
\][/tex]
Moles of H₂O ≈ 2.07 moles
4. Stoichiometry of the Reaction:
From the equation, 2 moles of H₂O are produced per reaction cycle, and this corresponds to the formation of 3 moles of sulfur (S). Therefore, the moles of sulfur formed are:
[tex]\[
\text{moles of S} = \left(\frac{3 \text{ moles of S}}{2 \text{ moles of H}_2\text{O}}\right) \times \text{moles of H}_2\text{O}
\][/tex]
Moles of S ≈ 3.11 moles
5. Mass of Sulfur:
Finally, we convert the moles of sulfur to grams:
[tex]\[
\text{mass of S} = \text{moles of S} \times \text{molar mass of S} = 3.11 \text{ moles} \times 32.07 \text{ g/mol}
\][/tex]
Mass of sulfur ≈ 99.8 g
Therefore, approximately 99.8 grams of sulfur are formed when 37.4 grams of water are produced, which matches the result in the options given. The correct answer is 99.8 g S.
1. Equation Analysis:
The balanced equation is:
[tex]\[
2 \text{H}_2\text{S(g)} + \text{SO}_2\text{(g)} \rightarrow 3 \text{S(s)} + 2 \text{H}_2\text{O(l)}
\][/tex]
2. Molar Masses:
- Molar mass of water (H₂O) ≈ 18.02 g/mol
- Molar mass of sulfur (S) ≈ 32.07 g/mol
3. Moles of Water:
First, calculate the moles of water formed:
[tex]\[
\text{moles of H}_2\text{O} = \frac{\text{mass of H}_2\text{O}}{\text{molar mass of H}_2\text{O}} = \frac{37.4 \text{ g}}{18.02 \text{ g/mol}}
\][/tex]
Moles of H₂O ≈ 2.07 moles
4. Stoichiometry of the Reaction:
From the equation, 2 moles of H₂O are produced per reaction cycle, and this corresponds to the formation of 3 moles of sulfur (S). Therefore, the moles of sulfur formed are:
[tex]\[
\text{moles of S} = \left(\frac{3 \text{ moles of S}}{2 \text{ moles of H}_2\text{O}}\right) \times \text{moles of H}_2\text{O}
\][/tex]
Moles of S ≈ 3.11 moles
5. Mass of Sulfur:
Finally, we convert the moles of sulfur to grams:
[tex]\[
\text{mass of S} = \text{moles of S} \times \text{molar mass of S} = 3.11 \text{ moles} \times 32.07 \text{ g/mol}
\][/tex]
Mass of sulfur ≈ 99.8 g
Therefore, approximately 99.8 grams of sulfur are formed when 37.4 grams of water are produced, which matches the result in the options given. The correct answer is 99.8 g S.
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