Question

What is the maximum mass of S8 that can be produced by combining 90.0 g of each reactant?
8SO2+16H2S>>>>>> 3S8+16H2O

Answer 1

Answer on Question #63569 - Chemistry - Other

Task:

What is the maximum mass of $S_8$ that can be produced by combining 90.0 g of each reactant?

$8SO_2 + 16H_2S \rightleftharpoons 3S_8 + 16H_2O$

Solution:

We can convert the masses of $SO_2$ and $H_2S$ to moles using molecular weights:

$$\begin{array}{l} \text{moles of } SO_2 = 90.0\,g \times \frac{1\,\text{mol } SO_2}{64.054\,g SO_2} = 1.405\,\text{mol } SO_2; \\ \text{moles of } H_2S = 90.0\,g \times \frac{1\,\text{mol } H_2S}{34.082\,g H_2S} = 2.6407\,\text{mol } H_2S; \end{array}$$

We calculate the actual molar ratio of the reactants, and then compare the actual ratio to the stoichiometric ratio from the balanced reaction.

$$\text{Actual ratio} = \frac{\text{moles of } H_2S}{\text{moles of } SO_2} = \frac{2.6407\,\text{mol } H_2S}{1.405\,\text{mol } SO_2} = \frac{1.8795\,\text{mol } H_2S}{1\,\text{mol } SO_2}$$

The actual ratio tells us that we have 1.8795 mol of $H_2S$ for every 1 mol of $SO_2$. In comparison, the stoichiometric ratio from our balanced reaction is below:

$$\text{Stoichiometric ratio} = \frac{16\,\text{mol } H_2S}{8\,\text{mol } SO_2} = \frac{2\,\text{mol } H_2S}{1\,\text{mol } SO_2}$$

This means we need at least 2 moles of $H_2S$ for every mole of $SO_2$. Since our actual ratio is smaller than our stoichiometric ratio, we have less $H_2S$ than we need to react with each mole of $SO_2$. Therefore, $H_2S$ is our limiting reagent and $SO_2$ is in excess.

Then,

$$\max_{\text{imum mass of } S_8} = \frac{3\,\text{mol } S_8 \times 256.472\,g S_8 \times 90.0\,g H_2S \times 1\,\text{mol } H_2S}{1\,\text{mol } S_8 \times 16\,\text{mol } H_2S \times 34.082\,g H_2S} = 126.987\,g S_8 \approx 127\,g S_8.$$

Answer: 127 g is the maximum mass of $S_8$.

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