Answer
2.1 L
Work Step by Step
Number of moles of $H_{2}$ to be produced=$ \frac{Mass\,in\,grams}{Molar\,mass}=\frac{25\,g}{2.01588\,g/mol}= 12.4\,mol$
It is clearly evident from the reaction that 3 moles of $H_{2}SO_{4}$ is needed to produce 3 moles of $H_{2}$.
This implies that 12.4 mol $H_{2}SO_{4}$ is needed to produce 12.4 mol $H_{2}.$
Volume of $H_{2}SO_{4}$ needed=$\frac{Number\,of\,moles\,needed}{Molarity}=\frac{12.4\,mol}{6.0\,M}=2.1\,L$
