Question

Magnesium hydroxide is the white milky substance in milk of magnesia. What mass of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ is formed when $15{\text{ mL}}$ of $0.2{\text{ M}} - {\text{NaOH}}$ is combined with $12{\text{ mL}}$ of $0.15{\text{ M}} - {\text{MgC}}{{\text{l}}_2}$?
A. $0.087{\text{ g}}$
B. $0.079{\text{ g}}$
C. $0.1044{\text{ g}}$
D. $0.522{\text{ g}}$

Answer 1

First write the balanced chemical equation for the reaction of ${\text{NaOH}}$ with ${\text{MgC}}{{\text{l}}_2}$. Then calculate the number of moles of both ${\text{NaOH}}$ and ${\text{MgC}}{{\text{l}}_2}$ and find the limiting reagent. From the limiting reagent calculate the number of moles and thus, the mass of magnesium hydroxide.

Complete answer:
First we have to write the balanced chemical equation for the reaction of ${\text{NaOH}}$ with ${\text{MgC}}{{\text{l}}_2}$.
When ${\text{NaOH}}$ reacts with ${\text{MgC}}{{\text{l}}_2}$, magnesium hydroxide and sodium chloride are formed. The balanced chemical equation is as follows:
${\text{2NaOH}} + {\text{MgC}}{{\text{l}}_2} \to {\text{Mg}}{\left( {{\text{OH}}} \right)_2} + {\text{NaCl}}$
Calculate the number of moles of ${\text{NaOH}}$ in $15{\text{ mL}}$ of $0.2{\text{ M}} - {\text{NaOH}}$ as follows:
Number of moles of ${\text{NaOH}}$ $= \dfrac{{0.2 \times 15}}{{1000}}$
Number of moles of ${\text{NaOH}}$ $= 0.003{\text{ mol}}$
Thus, the number of moles of ${\text{NaOH}}$ in $15{\text{ mL}}$ of $0.2{\text{ M}} - {\text{NaOH}}$ are $0.003{\text{ mol}}$.
Calculate the number of moles of ${\text{MgC}}{{\text{l}}_2}$ in $12{\text{ mL}}$ of $0.15{\text{ M}} - {\text{MgC}}{{\text{l}}_2}$ as follows:
Number of moles of ${\text{MgC}}{{\text{l}}_2}$ = $\dfrac{{0.15 \times 12}}{{1000}}$
Number of moles of ${\text{MgC}}{{\text{l}}_2}$ = $0.0018{\text{ mol}}$
Thus, the number of moles of ${\text{MgC}}{{\text{l}}_2}$ in $12{\text{ mL}}$ of $0.15{\text{ M}} - {\text{MgC}}{{\text{l}}_2}$ are $0.0018{\text{ mol}}$.
From the balanced chemical equation, we can see that one mole of ${\text{MgC}}{{\text{l}}_2}$ we require two moles of ${\text{NaOH}}$. Thus, now we will calculate the number of moles of ${\text{NaOH}}$ required for $0.0018{\text{ mol}}$ of ${\text{MgC}}{{\text{l}}_2}$. Thus,
$1{\text{ mol MgC}}{{\text{l}}_2} = {\text{2 mol NaOH}}$
Thus,
Number of moles of ${\text{NaOH}}$ required for $0.0018{\text{ mol}}$ of ${\text{MgC}}{{\text{l}}_2}$ $= \dfrac{{0.0018{\text{ mol MgC}}{{\text{l}}_2} \times 2{\text{ mol NaOH}}}}{{1{\text{ mol MgC}}{{\text{l}}_2}}}$
Number of moles of ${\text{NaOH}}$ required for $0.0018{\text{ mol}}$ of ${\text{MgC}}{{\text{l}}_2}$ $= 0.0036{\text{ mol NaOH}}$
Thus, the number of moles of ${\text{NaOH}}$ required for $0.0018{\text{ mol}}$ of ${\text{MgC}}{{\text{l}}_2}$ are $0.0036{\text{ mol}}$.
But we have only $0.003{\text{ mol}}$ of ${\text{NaOH}}$. Thus, ${\text{NaOH}}$ is the limiting reagent.
Now calculate the number of moles of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ formed using the number of moles of ${\text{NaOH}}$.
From the reaction stoichiometry, we can see that,
$1{\text{ mol Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}} = {\text{2 mol NaOH}}$
Thus,
Number of moles of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ $= \dfrac{{0.003{\text{ mol NaOH}} \times {\text{1 mol Mg}}{{\left( {{\text{OH}}} \right)}_2}}}{{2{\text{ mol NaOH}}}}$
Number of moles of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ $= 0.0015{\text{ mol Mg}}{\left( {{\text{OH}}} \right)_2}$
Thus, the number of moles of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ formed are $0.0015{\text{ mol}}$.

Now calculate the mass of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ in $0.0015{\text{ mol}}$ of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ as follows:
We know that,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Thus,
${\text{Mass}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molar mass}}\left( {{\text{g/mol}}} \right)$
Substitute $0.0015{\text{ mol}}$ for the number of moles of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ and $58.31{\text{ g/mol}}$ for the molar mass of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$. Thus,
${\text{Mass of Mg}}{\left( {{\text{OH}}} \right)_2} = 0.0015{\text{ mol}} \times 58.31{\text{ g/mol}}$
${\text{Mass of Mg}}{\left( {{\text{OH}}} \right)_2} = 0.087{\text{ g}}$
Thus, the mass of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ formed is $0.087{\text{ g}}$.

**Thus, the correct option is (A) $0.087{\text{ g}}$.

Note:**
While solving the problem, remember to write the correct balanced chemical equation for the reaction. A reagent that is completely used up in the reaction is known as the limiting reagent. The limiting reagent determines when the reaction stops. The amount of product formed in the reaction is limited by the limiting reagent.