Answer
2.1 L of that $6.0 \space M$ $H_2SO_4$ solution
Work Step by Step
1. Calculate the amount of moles of $ H_2 $:
$ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol
$$ 25.0 \space g \space H_2 \times \frac{1 \space mol \space H_2 }{ 2.016 \space g \space H_2 } = 12.4\underline{01} \space mol \space H_2 $$
2. Find the amount of moles of $ H_2SO_4 $ necessary:
$$ 12.4\underline{01} \times \frac{ 3 \space mol \space H_2SO_4 }{ 3 \space mol \space H_2 }= 12.4\underline{01} \space mol \space H_2SO_4 $$
3. Find the volume of the reactant:
$$ 12.4\underline{01} \space mol \space H_2SO_4 \times \frac{1 \space L \space H_2SO_4 }{ 6.0 \space mol \space H_2SO_4 } = 2.1 \space L \space H_2SO_4 $$
