# Question #3320d

Oct 11, 2016

120g ${S}_{8}$

#### Explanation:

The equation of the given reaction is

$8 S {O}_{2} + 16 {H}_{2} S \to 3 {S}_{8} + 16 {H}_{2} O$

Considering the atomic masses of

$S \to 32 \text{ g/mol}$

$O \to 16 \text{ g/mol}$

$H \to 1 \text{ g/mol}$

We calculate the molar masses of

$S {O}_{2} \to 32 + 2 \cdot 16 = 64 \text{ g/mol}$

${H}_{2} S \to 2 \cdot 1 + 32 = 34 \text{ g/mol}$

${S}_{8} \to 8 \cdot 32 = 256 \text{ g/mol}$

From the equation of the reaction we get the ratio of no. of moles of

${n}_{S {O}_{2}} : {n}_{{H}_{2} S} : {n}_{{S}_{8}} = 8 : 16 : 3$

So the ratio of their masses

$= {m}_{S {O}_{2}} : {m}_{{H}_{2} S} : {m}_{{S}_{8}}$

$= \left(8 \times 64\right) : \left(16 \times 34\right) : \left(3 \times 256\right)$

$= 16 : 17 : 24 = 80 : 85 : 120$

The mass of each reactant is 85g .

The ratio of masses as shown above reveals that 85g of ${H}_{2} S$ is exhausted when 80g of $S {O}_{2}$ reacts with it and as a result maximum 120g ${S}_{8}$ is produced.