Question #171b9

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Question #171b9

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Answer 1 · 2015-11-24T08:25:05

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

Explanation:

$2 H C l (a q) + M {g (O H)}_{2} (s) \to M g C l_{2} (a q) + 2 H_{2} O (l)$ $2HCl(aq) + Mg(OH)_2(s) rarr MgCl_2(aq) + 2H_2O(l)$

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 $35.75$ $35.75$ $m L$ $mL$ , $0.175$ $0.175$ $m o l \cdot L^{- 1}$ $mol*L^-1$ hydrochloric acid were used, then this represents a molar quantity of $35.75xx10^(-3)cancelLxx0.175$ $mol*cancelL^-1$ $=$ $6.26xx10^(-3) mol$ $HCl$ .

By the stoichiometry of the reaction, there were thus $3.23xx10^(-3) mol$ $Mg(OH)_2$ .

We had $0.125$ $mol*L^-1$ magnesium hydroxide solution available; thus $(3.23xx10^-3 cancel(mol))/(0.125*cancel(mol)*L^-1)$ gives an answer in litres (why?).

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Question #171b9

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