Question

Nitric acid can be neutralized by any base to form a salt and water, as in the following equation: $Mg(OH{)}_2+2HN{O}_3\to Mg(N{O}_3{)}_2+2H_{2}O$
What amount of magnesium nitrate salt will be formed by the reaction of $250\text{g}$ of $Mg(OH{)}_2$ and $250\text{g}$ of $HN{O}_3$?

• A. $294\text{g}$
• B. $147\text{g}$
• C. $636\text{g}$
• D. $318\text{g}$

Answer 1

The correct option is A $294\text{g}$

Moles of $Mg(OH{)}_2=\frac{250}{58}=4.31$ mol
Moles of $HN{O}_3=\frac{250}{63}=3.97$ mol
According to the reaction, $1$ mole of $Mg(OH{)}_2$ reacts with $2$ moles of $HN{O}_3$
$4.31$ moles of $Mg(OH{)}_2$ will react with $8.62$ moles of $HN{O}_3$
So $HN{O}_3$ is the limiting reagent.
Now, $2$ moles of $HN{O}_3$ produce $1$ mole of $Mg(N{O}_3{)}_2$\

$3.97$ moles of $HN{O}_3$ will produce $\frac{1}{2}\times 3.97=1.985$ mole of $Mg(N{O}_3{)}_2$
Mass of $1.985$ moles of $Mg(N{O}_3{)}_2$ = $1.985\times$molar mass of $Mg(N{O}_3{)}_2=1.985mol\times 148\text{g/mol}$
Mass of $Mg(N{O}_3{)}_2=293.78\text{g}\approx 294\text{g}$