# How do you solve this stoichiometry question?

## Feb 13, 2018

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation that describes this neutralization reaction.

${\text{MgO"_ ((s)) + 2"HNO"_ (3(aq)) -> "Mg"("NO"_ 3)_ (2(aq)) + "H"_ 2"O}}_{\left(l\right)}$

The chemical equation tells you that you need $2$ moles of nitric acid in order to consume $1$ mole of magnesium oxide, so start by calculating how many moles of each reactant are present.

To find the number of moles of magnesium oxide present in the sample, use the compound's molar mass

2.75 color(red)(cancel(color(black)("g"))) * "1 mole MgO"/(40.3044color(red)(cancel(color(black)("g")))) = "0.06823 moles MgO"

Now, in order for the reaction to consume all the moles of magnesium oxide, you would need

0.06823 color(red)(cancel(color(black)("moles MgO"))) * "2 moles HNO"_3/(1color(red)(cancel(color(black)("mole MgO")))) = "0.1365 moles HNO"_3

To see if you have enough moles of nitric acid present in the solution, use its molarity and volume.

70.0 color(red)(cancel(color(black)("mL solution"))) * "2.40 moles HNO"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.1680 moles HNO"_3

Since you have more moles of nitric acid than you need to neutralize all the moles of magnesium oxide

overbrace("0.1680 moles HNO"_3)^(color(blue)("what you have")) " " > " " overbrace("0.1365 moles HNO"_3)^(color(blue)("what you need"))

you can say that nitric acid is in excess, i.e. magnesium oxide acts as the limiting reagent. This means that the magnesium oxide will be completely consumed in the reaction.

The reaction will also consume $0.01365$ moles of nitric acid, leaving you with

${\text{0.1680 moles " - " 0.1365 moles" = "0.03150 moles HNO}}_{3}$

At this point, you can safely say that the resulting solution will be acidic, i.e. the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$, will be higher than the concentration of hydroxide anions, ${\text{OH}}^{-}$.

Now, the problem doesn't specify whether or not the volume of the solution changes upon the addition of the magnesium oxide, so you can assume that it remains constant.

This means that the resulting solution will contain $0.03159$ moles of nitric acid in $\text{70.0 mL}$ of the solution.

As you know, nitric acid is a strong acid that ionizes completely to produce hydronium cations in a $1 : 1$ mole ratio, so every mole of nitric acid will produce $1$ mole of hydronium cations in the solution.

overbrace("HNO"_ (3(aq)))^(color(blue)("1 mole ionizes")) + "H"_ 2"O"_ ((l)) -> overbrace("H"_ 3"O"_ ((aq))^(+))^(color(blue)("1 mole is produced")) + "NO"_ (3(aq))^(-)

This means that you have--remember to convert the volume of the solution to liters!

["H"_3"O"^(+)] = "0.03150 moles"/(70.0 * 10^(-3) quad "L") = color(darkgreen)(ul(color(black)("0.450 M")))

The answer is rounded to three sig figs.

Finally, you can use the fact that

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

to calculate the $\text{pH}$ of the solution, You will end up with

$\text{pH} = - \log \left(0.450\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{0.347}}}$

The answer is rounded to three decimal places, the number of sig figs you have for your values.