College

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are produced?

\[ 2 H_2 S(g) + SO_2(g) \rightarrow 3 S(s) + 2 H_2 O(l) \]

A) 56.1 g S
B) 44.4 g S
C) 14.0 g S
D) 99.8 g S
E) 66.6 g S

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

Other Questions