$$\ce{8H2(g) + S8(s) -> 8H2S(g)}$$

When $\pu{25.6 g}$ of $\ce{S8(s)}$ (molar mass $\pu{256 g mol^{-1}}$) reacts completely with an excess of $\ce{H2(g)}$ according to the equation above, the volume of $\ce{H2S(g)}$, measured at $0\degree$C and $\pu{1.00 atm}$, produced is closest to

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We can use the fact that the reaction occurs at STP to obtain the volume of gas produced.

$$\pu{25.6 g } \ce{ S8} \cdot \frac{\pu{1 mol }\ce{S8}}{\pu{256 g }\ce{ S8}}\cdot \frac{\pu{8 mol }\ce{H2S}}{\pu{1 mol }\ce{S8}} \cdot\frac{\pu{22.4 L }\ce{ S8}}{\pu{1 mol }\ce{ S8}}$$ $$=25.6\cdot \frac{1}{256} \cdot 8 \cdot 22.4 \text{ L }\ce{S8}$$ $$\approx \boxed{\pu{20 L }\ce{S8}}$$