# What volume of 0.20M $H_2SO_4$ is required to produce 34.0g of $H_2S$ by the reaction: $8KI+5H_2SO_4\rightarrow 4K_2SO_4+4I_2+H_2S+4H_2O$

$(a)\;24l\qquad(b)\;25l\qquad(c)\;28l\qquad(d)\;30l$

1 mole of $H_2S$=5 mole of $H_2SO_4$
$\large\frac{34}{34}$$=1$
1 mole of $H_2S$=5 mole of $H_2SO_4$
$0.20\times V=5$
$V=\large\frac{5}{0.20}$
$\Rightarrow 25l$