# Question ee6f7

Nov 20, 2015

$\text{0.714 M}$

#### Explanation:

All you have to do here is examine the mole ratio that exists between hydrochloric acid, $\text{HCl}$, and calcium carbonate, ${\text{CaCO}}_{3}$, and use it to determine how many moles of the former were needed to react with your calcium carbonate sample.

$\textcolor{red}{2} {\text{HCl"_text((aq]) + "CaCO"_text(3(aq]) ->"CaCl"_text(2(aq]) + "H"_2"O"_text((l]) + "CO}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrochloric acid and calcium carbonate. This tells you that the reaction will always consume $\textcolor{red}{2}$ time more moles of hydrochloric acid than of calcium carbonate.

So basically all you have to do here is figure out how many moles of calcium carbonate you have in that $\text{750 mg}$ sample.

To do that, use the compound's molar mass, which tells you what the exact mass of one mole of that substance is.

750 color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(1000color(red)(cancel(color(black)("mg")))) * ("1 mole CaCO"_3)/(100.09color(red)(cancel(color(black)("g")))) = "0.007493 moles CaCO"_3

This means that the hydrochloric acid solution must have contained

0.007943color(red)(cancel(color(black)("moles CaCO"_3))) * (color(red)(2)" moles HCl")/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "0.01589 moles HCl"

The molarity of the hdyrochloric acid solution will now be determined using the number of moles and the volume of the sample used in the titration - remember, th evolume must be expressed in liters!

$\textcolor{b l u e}{c = \frac{n}{V}}$

c = "0.01589 moles"/(22.25 * 10^(-3)"L") = color(green)("0.714 M")#

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of calcium carbonate.