Part A If 7.2 g of $$ \mathrm{CH}_{4} $$ and 6.5 g of S react, what is the mass, in grams, of $$ \mathrm{H}_{2} \mathrm{~S} $$ that is produced? $$ \mathrm{CH}_{4}(\mathrm{~g})+4 \mathrm{~S}(\mathrm{~g}) \rightarrow \mathrm{CS}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ Express your answer with the appropriate units. $$ \mathrm{m}\left(\mathrm{H}_{2} \mathrm{~S}\right)= $$ Value

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1. **Calculate the moles of $$ \mathrm{CH}_{4} $$:**

The molar mass of $$ \mathrm{CH}_{4} $$ is approximately  
   $$
   12.0 + 4(1.0) = 16.0\ \mathrm{g/mol}
   $$
   so the moles of $$ \mathrm{CH}_{4} $$ are  
   $$
   \text{moles of } \mathrm{CH}_{4} = \frac{7.2\ \mathrm{g}}{16.0\ \mathrm{g/mol}} = 0.45\ \mathrm{mol}.
   $$

2. **Calculate the moles of $$ \mathrm{S} $$:**

The molar mass of $$ \mathrm{S} $$ is approximately  
   $$
   32.0\ \mathrm{g/mol}
   $$
   so the moles of $$ \mathrm{S} $$ are  
   $$
   \text{moles of } \mathrm{S} = \frac{6.5\ \mathrm{g}}{32.0\ \mathrm{g/mol}} \approx 0.203\ \mathrm{mol}.
   $$

3. **Determine the limiting reagent:**

The balanced reaction is  
   $$
   \mathrm{CH}_{4} + 4\ \mathrm{S} \rightarrow \mathrm{CS}_{2} + 2\ \mathrm{H}_{2}\mathrm{S}.
   $$
   From the stoichiometry, 1 mole of $$ \mathrm{CH}_{4} $$ requires 4 moles of $$ \mathrm{S} $$.

- For $$ 0.45\ \mathrm{mol} $$ $$ \mathrm{CH}_{4} $$, the required moles of $$ \mathrm{S} $$ would be:
     $$
     0.45\ \mathrm{mol} \times 4 = 1.8\ \mathrm{mol}.
     $$
     
   Since only $$ \approx 0.203\ \mathrm{mol} $$ of $$ \mathrm{S} $$ is available, $$ \mathrm{S} $$ is the limiting reagent.

4. **Calculate the moles of $$ \mathrm{H}_{2}\mathrm{S} $$ produced:**

From the balanced equation, 4 moles of $$ \mathrm{S} $$ produce 2 moles of $$ \mathrm{H}_{2}\mathrm{S} $$. Thus, the mole ratio is:
   $$
   \frac{2\ \mathrm{mol}\ \mathrm{H}_{2}\mathrm{S}}{4\ \mathrm{mol}\ \mathrm{S}} = 0.5.
   $$
   The moles of $$ \mathrm{H}_{2}\mathrm{S} $$ produced are then:
   $$
   0.203\ \mathrm{mol}\ \mathrm{S} \times 0.5 = 0.1015\ \mathrm{mol}\ \mathrm{H}_{2}\mathrm{S}.
   $$

5. **Convert moles of $$ \mathrm{H}_{2}\mathrm{S} $$ to grams:**

The molar mass of $$ \mathrm{H}_{2}\mathrm{S} $$ is approximately  
   $$
   2(1.0) + 32.0 = 34.0\ \mathrm{g/mol}.
   $$
   Thus, the mass of $$ \mathrm{H}_{2}\mathrm{S} $$ produced is:
   $$
   0.1015\ \mathrm{mol} \times 34.0\ \mathrm{g/mol} \approx 3.45\ \mathrm{g}.
   $$

$$
\mathrm{m}\left(\mathrm{H}_{2}\mathrm{S}\right)=3.45\ \mathrm{g}
$$
The mass of $$ \mathrm{H}_{2}\mathrm{S} $$ produced is 3.45 grams.

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