Answer :
When 35.8 g of [tex]\( \text{CH}_4 \)[/tex] and 75.5 g of [tex]\( \text{S} \)[/tex] react, approximately 40.06 g of [tex]\( \text{H}_2\text{S} \)[/tex] are produced. This calculation involved determining the limiting reactant, using stoichiometry, and converting between moles and grams to find the mass of the product.
To determine how many grams of [tex]\( \text{H}_2\text{S} \)[/tex] are produced when 35.8 g of [tex]\( \text{CH}_4 \)[/tex] (methane) and 75.5 g of [tex]\( \text{S} \)[/tex] (sulfur) react, we need to first write and balance the chemical equation for the reaction:
[tex]\[ \text{CH}_4 + 2\text{S} \rightarrow \text{CS}_2 + 2\text{H}_2\text{S} \][/tex]
However, it is important to use the correct stoichiometry for the given reaction. Considering the reaction between methane and sulfur can produce carbon disulfide [tex](\( \text{CS}_2 \))[/tex] and hydrogen sulfide [tex](\( \text{H}_2\text{S} \))[/tex], the balanced chemical equation should be:
[tex]\[ \text{CH}_4 + 4\text{S} \rightarrow \text{CS}_2 + 2\text{H}_2\text{S} \][/tex]
Next, we need to determine the molar masses of the reactants and the product:
- Molar mass of [tex]\( \text{CH}_4 \): \( 12.01 + 4 \times 1.01 = 16.05 \, \text{g/mol} \)[/tex]
- Molar mass of [tex]\( \text{S} \): \( 32.07 \, \text{g/mol} \)[/tex]
- Molar mass of [tex]\( \text{H}_2\text{S} \): \( 2 \times 1.01 + 32.07 = 34.09 \, \text{g/mol} \)[/tex]
First, we calculate the number of moles of each reactant:
- Moles of [tex]\( \text{CH}_4 \):[/tex]
[tex]\[ \frac{35.8 \, \text{g}}{16.05 \, \text{g/mol}} \approx 2.23 \, \text{mol} \][/tex]
- Moles of [tex]\( \text{S} \):[/tex]
[tex]\[ \frac{75.5 \, \text{g}}{32.07 \, \text{g/mol}} \approx 2.35 \, \text{mol} \][/tex]
Now, using the balanced equation, we see that 1 mole of [tex]\( \text{CH}_4 \)[/tex] reacts with 4 moles of [tex]\( \text{S} \)[/tex] to produce 2 moles of [tex]\( \text{H}_2\text{S} \).[/tex] Therefore, we need to determine the limiting reactant.
- For 2.23 moles of [tex]\( \text{CH}_4 \)[/tex], we need:
[tex]\[ 2.23 \, \text{mol} \times 4 \, \frac{\text{mol S}}{\text{mol CH}_4} = 8.92 \, \text{mol S} \][/tex]
We only have 2.35 moles of [tex]\( \text{S} \)[/tex], which is less than the 8.92 moles required, making [tex]\( \text{S} \)[/tex] the limiting reactant.
Since [tex]\( \text{S} \)[/tex] is the limiting reactant, we use its quantity to determine the amount of [tex]\( \text{H}_2\text{S} \)[/tex] produced. According to the balanced equation, 4 moles of [tex]\( \text{S} \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{S} \)[/tex]:
- Moles of [tex]\( \text{H}_2\text{S} \)[/tex] produced:
[tex]\[ 2.35 \, \text{mol S} \times \frac{2 \, \text{mol H}_2\text{S}}{4 \, \text{mol S}} = 1.175 \, \text{mol H}_2\text{S} \][/tex]
Finally, we convert the moles of [tex]\( \text{H}_2\text{S} \)[/tex] to grams:
[tex]\[ 1.175 \, \text{mol} \times 34.09 \, \text{g/mol} \approx 40.06 \, \text{g} \][/tex]
Therefore, when 35.8 g of [tex]\( \text{CH}_4 \)[/tex] and 75.5 g of [tex]\( \text{S} \)[/tex] react, approximately 40.06 g of [tex]\( \text{H}_2\text{S} \)[/tex] are produced. This calculation involved determining the limiting reactant, using stoichiometry, and converting between moles and grams to find the mass of the product.
