# If 1g sample of limestone was dissolved in acid to give 0.38 g of "CO"_2. If limestone contains no carbonate other than "CaCO"_3, find the percentage of "CaCO"_3 in the limestone?

## $\text{CaCO"_3(s) + 2"HCl"(aq) -> "CO"2(g)+ "CaCl"_2(aq) + "H"_2"O} \left(l\right)$

Apr 25, 2017

${\text{86% CaCO}}_{3}$

#### Explanation:

Notice that the problem provides you with grams of carbon dioxide, but that the balanced chemical equation uses moles, so right from the start, you know that you must convert grams to moles or vice versa.

${\text{CaCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)} \uparrow$

You know that $1$ mole of calcium carbonate reacts to produce $1$ mole of carbon dioxide. Use the molar masses of the two compounds to convert this mole ratio to a gram ratio.

You know that

M_ ("M CaCO"_ 3) = "100.1 g mol"^(-1)

M_ ("M CO"_ 2) = "44.01 g mol"^(-1)

This means that when $\text{100.1 g}$ of calcium carbonate take part in the reaction, the equivalent of $1$ mole of calcium carbonate, the reaction produces $\text{44.01 g}$ of carbon dioxide, the equivalent of $1$ mole of carbon dioxide.

You thus have

overbrace("1 mole CaCO"_3/"1 mole CO"_2)^(color(blue)("mole ratio")) = overbrace("100.1 g CaCO"_3/"44.01 g CO"_2)^(color(purple)("gram ratio"))

Use the gram ratio to determine how many grams of calcium carbonate were needed to produce $\text{0.38 g}$ of carbon dioxide

0.38 color(red)(cancel(color(black)("g CO"_2))) * "100.1 g CaCO"_3/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "0.864 g CaCO"_3

To find the percentage of calcium carbonate in the sample of limestone, simply divide the mass of calcium carbonate by the total mass of the limestone and multiply the result by 100%

"% CaCO"_3 = (0.864 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(86%)))

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of limestone.