# In the equation Mg + 2HCl -> MgCl_2 + H_2, what volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75 g of HCl?

Feb 28, 2016

$1.03$ mol of dihydrogen gas will evolve, with a volume slightly over $22.4$ ${\mathrm{dm}}^{3}$ at $S T P$.

#### Explanation:

Moles of magnesium: $\frac{50.0 \cdot g}{24.31 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.06$ $m o l$

Moles of hydrogen chloride gas: $\frac{75.0 \cdot g}{36.2 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.07$ $m o l$

Given equimolar amounts, clearly, the gas is in deficiency as the stoichiometric equation requires 2 equiv of gas per equiv of metal.

So, the $1$ mole of gas evolves, (conveniently, as at $S T P$ the molar volume is $22.4$ ${\mathrm{dm}}^{3}$).