# Question 57ffe

Jun 14, 2015

The concentration of the nitric acid solution was 5.6 M.

#### Explanation:

Start by writing the balanced chemical equation for your reaction

$\textcolor{red}{2} H N {O}_{3 \left(a q\right)} + M g C {O}_{3 \left(s\right)} \to M g {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} + C {O}_{2 \left(g\right)}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between nitric acid and magnesium carbonate. This tells you that, regardless of how many moles of magnesium carbonate react, the reaction will always use 2 times more moles of nitric acid.

Use magnesium carbonate's molar mass to determine how many moles were present in the 8.2-g sample that completely reacted

8.2cancel("g") * "1 mole"/(84.31cancel("g")) = "0.09726 moles" $M g C {O}_{3}$

This means that the nitric acid solution had

0.09726cancel("moles"MgCO_3) * (color(red)(2)" moles"HNO_3)/(1cancel("mole"MgCO_3)) = "0.1945 moles"# $H N {O}_{3}$

Molarity is defined as moles of solute, in your case nitric acid, divided by liters of solution. Since you know the volume of the nitric acid solution and how many moles of solute it contained, you can solve for its molarity by

$C = \frac{n}{V} = \text{0.1945 moles"/(35.0 * 10^(-3)"L") = 5.557 "moles"/"L}$

Rounded to two sig figs, the number of sig figs you gave for the mass of magnesium carbonate, the answer will be

$C = \textcolor{g r e e n}{\text{5.6 M}}$