# Question c46be

Dec 20, 2015

Here's what I got.

#### Explanation:

Provided that those are the values given to you, your answer seems correct to me. I'll go through the problem just to make sure that we get everything right.

Start by taking a look at the balanced chemical equation for this reaction

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

Two important things to notice here

• you have a $\textcolor{red}{2} : 1$ mole ratio between hydrochloric acid and calcium carbonate
• you have a $1 : 1$ mole ratio between calcium carbonate and carbon dioxide

Use the molarity of the hydrochloric acid solution to determine how many moles of acid are present

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

${n}_{H C l} = \text{6 M" * 50 * 10^(-3)"L" = "0.30 moles HCl}$

Now use calcium carbonate's molar mass to determine how many moles you have in that $\text{10-g}$ sample

10 color(red)(cancel(color(black)("g"))) * ("1 mole CaCO"_3)/(100.1 color(red)(cancel(color(black)("g")))) ~~ "0.10 moles CaCO"_3

Notice that you have more hydrochloric acid that you would need, since according to the $\textcolor{red}{2} : 1$ mole ratio that exists between the two compounds, you'd only need

0.10 color(red)(cancel(color(black)("moles CaCO"_3))) * (color(red)(2)" moles HCl")/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "0.20 moles HCl"

This means that calcium carbonate acts as a limiting reagent, i.e. all the moles of calcium carbonate will be consumed by the reaction.

Implicitly, the reaction will produce

0.10 color(red)(cancel(color(black)("moles CaCO"_3))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "0.10 moles CO"_2

If you use the fact that the molar volume of a gas at STP is given to you as $\text{22.4 L}$, then the volume of carbon dioxide produced by the reaction will be

0.10 color(red)(cancel(color(black)("moles"))) * "22.4 L"/(1color(red)(cancel(color(black)("mole")))) = "2.24 L"

Let's work backwards from the solution given by the textbook and see what would you need in order to get that answer.

So, if the volume of carbon dioxide is $\text{6.72 L}$, then you can say that the reaction produced

6.72 color(red)(cancel(color(black)("L"))) * "1 mole CO"_2/(22.4color(red)(cancel(color(black)("L")))) = "0.30 moles CO"_2#

This means that the reaction must have consumed $0.30$ moles of calcium carbonate. This corresponds to an initial $\text{30-g}$ sample of ${\text{CaCO}}_{3}$, in the context of the compound still acting as a limiting reagent.

In turn, this implies that at least $0.60$ moles of hydrochloric acid were present in solution.

So, as a conclusion, either the answer given by the book is wrong, or the values given in the problem are wrong. Either way, your calculations are correct.