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Answer :
To find out how many grams of sulfur are formed when 37.4 grams of water are produced, we can use the stoichiometry of the chemical reaction:
[tex]\[ 2 \, \text{H}_2\text{S} (g) + \text{SO}_2 (g) \rightarrow 3 \, \text{S} (s) + 2 \, \text{H}_2\text{O} (l) \][/tex]
Step 1: Calculate the moles of water
First, we need to determine the number of moles of water (H₂O) formed. The molar mass of water is approximately 18.015 grams per mole.
[tex]\[
\text{Moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} = \frac{37.4 \, \text{g}}{18.015 \, \text{g/mol}} \approx 2.076 \, \text{moles}
\][/tex]
Step 2: Use stoichiometry to find the moles of sulfur
From the balanced equation, we see that 2 moles of H₂O are produced for every 3 moles of sulfur (S) formed. Therefore, the moles of sulfur that are formed are calculated as follows:
[tex]\[
\text{Moles of S} = \left(\frac{3 \, \text{moles of S}}{2 \, \text{moles of H}_2\text{O}}\right) \times 2.076 \, \text{moles of H}_2\text{O} \approx 3.114 \, \text{moles}
\][/tex]
Step 3: Calculate the mass of sulfur
Finally, we convert the moles of sulfur to grams using the molar mass of sulfur, which is approximately 32.06 grams per mole:
[tex]\[
\text{Mass of S} = \text{Moles of S} \times \text{Molar mass of S} = 3.114 \, \text{moles} \times 32.06 \, \text{g/mol} \approx 99.8 \, \text{g}
\][/tex]
So, the mass of sulfur formed is approximately 99.8 grams. Therefore, the correct answer is:
D) 99.8 g S
[tex]\[ 2 \, \text{H}_2\text{S} (g) + \text{SO}_2 (g) \rightarrow 3 \, \text{S} (s) + 2 \, \text{H}_2\text{O} (l) \][/tex]
Step 1: Calculate the moles of water
First, we need to determine the number of moles of water (H₂O) formed. The molar mass of water is approximately 18.015 grams per mole.
[tex]\[
\text{Moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} = \frac{37.4 \, \text{g}}{18.015 \, \text{g/mol}} \approx 2.076 \, \text{moles}
\][/tex]
Step 2: Use stoichiometry to find the moles of sulfur
From the balanced equation, we see that 2 moles of H₂O are produced for every 3 moles of sulfur (S) formed. Therefore, the moles of sulfur that are formed are calculated as follows:
[tex]\[
\text{Moles of S} = \left(\frac{3 \, \text{moles of S}}{2 \, \text{moles of H}_2\text{O}}\right) \times 2.076 \, \text{moles of H}_2\text{O} \approx 3.114 \, \text{moles}
\][/tex]
Step 3: Calculate the mass of sulfur
Finally, we convert the moles of sulfur to grams using the molar mass of sulfur, which is approximately 32.06 grams per mole:
[tex]\[
\text{Mass of S} = \text{Moles of S} \times \text{Molar mass of S} = 3.114 \, \text{moles} \times 32.06 \, \text{g/mol} \approx 99.8 \, \text{g}
\][/tex]
So, the mass of sulfur formed is approximately 99.8 grams. Therefore, the correct answer is:
D) 99.8 g S
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