Question #171b9

1 Answer
anor277
Nov 24, 2015

About 26 mL of the magnesium solution.

Explanation:

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 mL, 0.175 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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