# What volume of 6.0 M HCl is required to exactly react with 0.040g Mg?

Dec 16, 2016

We need, (i) a stoichiometrically balanced equation........

$M g \left(s\right) + 2 H C l \left(a q\right) \rightarrow M g C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

#### Explanation:

Given the equation, we need (ii) $\text{2 equiv}$ of acid per $\text{equiv}$ metal.

And thus, $\frac{0.040 \cdot g}{24.31 \cdot g \cdot m o {l}^{-} 1} \times 2 \times \frac{{10}^{3} \cdot m L \cdot {L}^{-} 1}{6.0 \cdot m o l \cdot {L}^{-} 1} = 0.55 \cdot m L$

This is not a large volume in that the acid is quite concentrated, and the metal is quite low atomic mass.