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 g of magnesium hydroxide and 250 g of nitric acid?

• A. 294 g
• B. 147 g
• C. 636 g
• D. 318 g

Answer 1

The correct option is A 294 g

Moles of $Mg(OH{)}_2=\frac{250}{58}=4.31$ moles
Moles of $HN{O}_3=\frac{250}{63}=3.97$ moles
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 $Mg(N{O}_3{)}_2=1.985\times 148$g
Mass of $Mg(N{O}_3{)}_2=293.78$ g ~ 294 g