Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters (L) produced at 300 K and 1.0 atm pressure, when 5.4 g of aluminium and 50.0 mL of 5.0 M sulfuric acid are combined for the reaction?
(Use molar mass of aluminium as $$27.0 g mol^{−1}, 𝑅 = 0.082 atm L mol^{−1} K^{−1}$$)

The balanced reaction of aluminium with sulphuric acid is
$$2\,Al + 3\,H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3\,H_2$$

Step 1 - Moles of each reactant
Molar mass of aluminium = $$27.0\;g\;mol^{-1}$$.
Moles of $$Al$$ present = $$\frac{5.4\;g}{27.0\;g\;mol^{-1}} = 0.20\;mol$$ $$-(1)$$

Volume of acid solution = $$50.0\;mL = 0.050\;L$$.
Molarity of acid = $$5.0\;M$$, so
Moles of $$H_2SO_4$$ present = $$0.050\;L \times 5.0\;mol\,L^{-1} = 0.25\;mol$$ $$-(2)$$

Step 2 - Identify the limiting reagent
From the balanced equation, $$2\;mol\;Al$$ need $$3\;mol\;H_2SO_4$$.
Required $$H_2SO_4$$ for $$0.20\;mol\;Al$$:
$$0.20 \times \frac{3}{2} = 0.30\;mol$$ $$-(3)$$
Only $$0.25\;mol$$ acid is available (see $$(2)$$), which is less than $$0.30\;mol$$, therefore $$H_2SO_4$$ is the limiting reagent.

Step 3 - Moles of hydrogen produced
Stoichiometry gives $$3\;mol\;H_2SO_4 \rightarrow 3\;mol\;H_2$$, i.e. a 1 : 1 ratio.
Thus moles of $$H_2$$ formed = moles of limiting $$H_2SO_4$$ = $$0.25\;mol$$ $$-(4)$$

Step 4 - Volume of hydrogen at 300 K and 1.0 atm
Using the ideal-gas equation $$PV = nRT$$,
$$V = \frac{nRT}{P} = \frac{0.25\;mol \times 0.082\;atm\,L\,mol^{-1}\,K^{-1} \times 300\;K}{1.0\;atm}$$

$$V = 0.25 \times 24.6 = 6.15\;L$$ $$-(5)$$

Therefore the volume of hydrogen gas produced is 6.15 L.