Question

What is the maximum mass of S8 that can be produced by combining 84.0 g of each reactant?

8SO2 + 16H2S ---> 3S8 + 16H2O

Answer 1

Answer on Question #40489, Chemistry, Other

**Task:**

What is the maximum mass of $S_8$ that can be produced by combining 84.0 g of each reactant?

$$8 \mathrm{SO_2} + 16 \mathrm{H_2S} = 3 \mathrm{S_8} + 16 \mathrm{H_2O}$$

**Answer:**

$$v = \frac{m}{M}$$

where m = mass, grams;

M = molar mass, gram/mol.

$$M(SO_2) = 64.1 \text{ g/mol} \quad M(H_2S) = 34.1 \text{ g/mol}$$ $$v(SO_2) = \frac{84.0}{64.1} = 1.31 \text{ moles}$$ $$v(H_2S) = \frac{84.0}{34.1} = 2.46 \text{ moles}$$

Let's calculate the amount of $S_8$, that can be produced from 84.0 grams of each reactant:

$$v(S_8) = \frac{v(SO_2)}{8} \cdot 3 = \frac{1.31}{8} \cdot 3 = 0.49 \text{ moles}$$ $$v(S_8) = \frac{v(H_2S)}{16} \cdot 3 = \frac{2.46}{16} \cdot 3 = 0.46 \text{ moles}$$

As we can see from the previous calculations, the amount of $H_2S$ is the determining factor.

There will be an excess amount of $SO_2$. That is why:

$$m(S_8) = v(S_8) \cdot M(S_8)$$ $$M(S_8) = 256.5 \text{ g/mol}$$

That is why the maximum mass of $S_8$, that can be produced is equal to:

$$m(S_8) = 0.46 \cdot 256.5 = 118 \text{ g}$$