# What volume of hydrogen gas is obtained when "43.39 g" of sulfuric acid reacts with solid aluminum metal?

May 13, 2017

The diligent chemist reads the question carefully and realizes that the bare bones of the reaction were already given to him:

"H"_2"SO"_4(aq) + "Al"(s) -> "H"_2(g) + ???

He wracks his brain for a moment, initially expecting that if a strong acid reacts with metal, it oxidizes the metal and produces ${\text{SO}}_{2} \left(g\right)$, or even $\text{H"_2"S} \left(g\right)$. The acid, he remarks, is likely dilute, and given excess metal, the acid is of course the limiting reactant.

He believes the following reaction is then reasonable:

$3 {\text{H"_2"SO"_4(aq) + 2"Al"(s) -> "Al"_2("SO"_4)_3(aq) + 3"H}}_{2} \left(g\right)$

Given the $\text{43.39 g}$ of ${\text{H"_2"SO}}_{4}$, he is too lazy to calculate the molar mass of sulfuric acid, so he performs a quick google search to locate the non-obscure quantity, $\text{98.079 g/mol}$.

He realizes the proper unit conversion is from $\text{g}$ to $\text{mol acid}$ to ${\text{mol H}}_{2} \left(g\right)$, but vehemently takes issue with the lack of information in the problem, and decides to assume a temperature of $\text{298.15 K}$ and pressure of $\text{1 atm}$ for the remainder of the problem.

$43.39 \cancel{{\text{g H"_2"SO"_4) xx cancel("1 mol H"_2"SO"_4)/(98.079 cancel("g H"_2"SO"_4)) xx "1 mol H"_2/cancel("1 mol H"_2"SO}}_{4}}$

$=$ ${\text{0.4424 mol H}}_{2}$

He proceeds to rapidly utilize the ideal gas law, as if it were a daily routine, to obtain:

$\textcolor{b l u e}{{V}_{{H}_{2} \left(g\right)}} = \frac{n R T}{P}$

= (("0.4424 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("1 atm")

$=$ $\textcolor{b l u e}{{\text{10.82 L H}}_{2}}$