Question

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

Answer 1

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 `"SO"_2(g)`, or even `"H"_2"S"(g)`. 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"H"_2"SO"_4(aq) + 2"Al"(s) -> "Al"_2("SO"_4)_3(aq) + 3"H"_2(g)`

Given the `"43.39 g"` of `"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, `"98.079 g/mol"`.

He realizes the proper unit conversion is from `"g"` to `"mol acid"` to `"mol H"_2(g)`, but vehemently takes issue with the lack of information in the problem, and decides to assume a temperature of `"298.15 K"` and pressure of `"1 atm"` for the remainder of the problem.

`43.39 cancel("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)`

`=` `"0.4424 mol H"_2`

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

`color(blue)(V_(H_2(g))) = (nRT)/P`

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

`=` `color(blue)("10.82 L H"_2)`