Thank you for visiting According to the following reaction how many grams of sulfur are formed when 37 4 g of water are formed tex 2 text H 2. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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 (Hâ‚‚O) ≈ 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 Hâ‚‚O ≈ 2.07 moles
4. Stoichiometry of the Reaction:
From the equation, 2 moles of Hâ‚‚O 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 (Hâ‚‚O) ≈ 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 Hâ‚‚O ≈ 2.07 moles
4. Stoichiometry of the Reaction:
From the equation, 2 moles of Hâ‚‚O 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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