Aluminium reacts with sulphuric acid to form aluminium sulfate and hydrogen.  What is the volume of hydrogen gas in litres (L) produced at 300 K and 1.0 atm pressure, when 8.1 g of aluminium and 50.0 mL of 5.0 M sulphuric acid are combined for the reaction? (Use molar mass of aluminium as ${\text{20.0 g mol}}^{\text{-1}}{\text{,R=0.082\hspace{0.17em}\hspace{0.17em}atm\hspace{0.17em}\hspace{0.17em}L\hspace{0.17em}\hspace{0.17em}mol}}^{\text{-1}}{\text{\hspace{0.17em}\hspace{0.17em}K}}^{\text{-1}}$)

Number of moles of aluminium $=\frac{8.1}{27}=0.3$

            Number of moles of $H_{2}S{O}_4=\frac{MV}{1000}=\frac{5\times 50}{1000}=0.25$

Case I: Let aluminium is fully consumed:

Number of moles of $H_2$ obtained from 0.2 mol of $Al$

$=\frac{3}{2}\times 0.2=0.3$

Case II: Let $H_{2}S{O}_4$ is fully consumed:

Number of moles of $H_2$ obtained from 0.25 mol $H_{2}S{O}_4$

$=\frac{3}{3}\times 0.25=0.25$

Since $H_{2}S{O}_4$ on complete consumption gives least amount of $H_2$, it will be limiting reactant and actual amount of $H_2$ formed will be 0.25 mol.  Volume of $H_2$ gas can be calculated as:

$\begin{array}{c}PV=nRT\\ V=\frac{nRT}{P}=\frac{0.25\times 0.082\times 300}{1}=6.15L\end{array}$