# Question 472ec

Mar 22, 2016

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation for this reaction

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

Notice that the reaction consumes $\textcolor{red}{2}$ moles of hydrochloric acid, $\text{HCl}$, for every $1$ mole of calcium carbonate, ${\text{CaCO}}_{3}$, that takes part in the reaction.

This means that if you know how many moles of calcium carbonate took part in the reaction, you can use this $\textcolor{red}{2} : 1$ mole ratio that exists between the two reactants to determine how many moles of hydrochloric acid were needed.

So, you know that you start with a $\text{5.00-g}$ sample of calcium carbonate. After the reaction is finished, you are left with $\text{2.00 g}$ of unreacted calcium carbonate.

This tells you that the reaction consumed only

${m}_{C a C {O}_{3}} = 5.00 - 2.00 = {\text{3.00 g CaCO}}_{3}$

Use calcium carbonate's molar mass to determine how many moles you'd get in this many grams

$3.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * "1 mole CaCO"_3/(100.09color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)"0.0300 moles CaCO}}_{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now that you know how many moles of calcium carbonate reacted, you can sue the aforementioned mole ratio to get the number of moles of hydrochloric acid that reacted

$0.0300 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CaCO"_3))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = color(green)(|bar(ul(color(white)(a/a)"0.0600 moles HCl} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finally, to get the molarity of the hydrochloric acid solution, use the fact that molarity is defined as moles of solute, which in this case is hydrochloric acid, divided by liters of solution

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Convert the volume of the solution from milliliters to liters by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 L" = 10^3"mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

c_(HCl) = "0.0600 moles"/(55.0 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"1.09 M"color(white)(a/a)|)))#

The answers are rounded to three sig figs.