# 21.10 grams of magnesium react with excess sulfuric acid. How many liters of hydrogen are produced?

Feb 22, 2016

We have (almost!) a mole of magnesium atoms. So we generate, almost a mole of dihydrogen gas, approx $25$ ${\mathrm{dm}}^{3}$!

#### Explanation:

$M g \left(s\right) + {H}_{2} S {O}_{4} \left(a q\right) \rightarrow M g S {O}_{4} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

Given the equation (which you should be able to reproduce) there is a 1:1 equivalence of magnesium metal to dihydrogen gas.

Moles of $M g$ $=$ $\frac{21.10 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$.

Now it is a fact, that at $1$ atmosphere pressure and $298$ $K$, $1$ $m o l$ of ideal gas (and we assume dihydrogen approximates ideality!) occupies $24.5$ ${\mathrm{dm}}^{3}$ ($1$ ${\mathrm{dm}}^{3}$ $=$ $1$ $L$).

So volume of dihyfrogen (at $S L C$) $=$

$\frac{21.10 \cdot \cancel{g}}{26.98 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}}$ $\times$ $24.5 \cdot L \cdot \cancel{m o {l}^{-} 1}$ $=$ ??L