# Question #171b9

Nov 24, 2015

About $26$ $m L$ of the magnesium solution.

#### Explanation:

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

The equation above is balanced, and gives a stoichiometric picture of the reactivity. Two equiv hydrochloric acid are required to neutralize the magnesium salt.

If $35.75$ $m L$, $0.175$ $m o l \cdot {L}^{-} 1$ hydrochloric acid were used, then this represents a molar quantity of $35.75 \times {10}^{- 3} \cancel{L} \times 0.175$ $m o l \cdot {\cancel{L}}^{-} 1$ $=$ $6.26 \times {10}^{- 3} m o l$ $H C l$.

By the stoichiometry of the reaction, there were thus $3.23 \times {10}^{- 3} m o l$ $M g {\left(O H\right)}_{2}$.

We had $0.125$ $m o l \cdot {L}^{-} 1$ magnesium hydroxide solution available; thus $\frac{3.23 \times {10}^{-} 3 \cancel{m o l}}{0.125 \cdot \cancel{m o l} \cdot {L}^{-} 1}$ gives an answer in litres (why?).