# Question ef8fc

Feb 12, 2016

$\text{37 g}$

#### Explanation:

Like always, start with a balanced chemical equation for this decomposition reaction

${\text{CaCO"_text(3(s]) stackrel(color(red)(Delta)color(white)(aa))(->) "CaO"_text((s]) + "CO}}_{\textrm{2 \left(g\right]}}$ $\uparrow$

Notice that you have $1 : 1$ mole ratios across the board. One mole of calcium carbonate will decompose to form one mole of calcium oxide and one mole of carbon dioxide.

Keep this in mind.

Now, since you didn't provide information about the temperature and pressure at which the oxygen gas is being collected, I"ll assume that you're at STP, Standard Pressure and Temperature.

Under STP condittions, which are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, one mole of any ideal gas is known to occupy $\text{22.7 L}$ - this is known as the molar volume of a gas at STP.

This means that you can use the volume of carbon dioxide to figure out how many moles of oxygen were produced by the reaction

6.72 color(red)(cancel(color(black)("L"))) * overbrace("1 mole O"_2/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas")) = "0.2960 moles CO"_2

Now go back to the mole ratio that exists between calcium carbonate and carbon dioxide. If the reaction produced $0.2960$ moles of carbon dioxide, then it must have consumed $0.2960$ moles of calcium carbonate.

Use calcium carbonate's molar mass to find how many grams of calcium carbonate would contain this many moles

0.2960 color(red)(cancel(color(black)("moles CaCO"_3))) * "100.1 g"/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "29.63 g"

Now, you know that your initial sample contained 20% impurities. This is equivalent to saying that it contained 80% calcium carbonate.

If that is the case, then the initial mass of the sample was

29.63 color(red)(cancel(color(black)("g CaCO"_3))) * overbrace("100 g sample"/(80color(red)(cancel(color(black)("g CaCO"_3)))))^(color(purple)(=80%"CaCO"_3)) = color(green)("37 g")#

I'll leave the answer rounded to two sig figs.

SIDE NOTE Make sure to use the value of the molar volume of a gas at STP given to you. Usually, that value is $\text{22.4 L}$, so if that's what you have, simply redo the calculations by replacing $\text{22.7 L}$ with your value.

Feb 12, 2016

If the volume of CO2 ie 6.72L is in STP, then calculation may be made as follows

#### Explanation:

According to the reaction CaCO3 (s)------> CaO(s) + CO2(g)
1 mole or 22.4 L CO2 at STP is obtained on thermal decomposition of 100 g of CaCO3,
So 6.72L CO2 can be obtained from 30 g CaCO3. As the sample contains 20% impurities,it can be said that 80 g CaCO3 (which will only produce CO2) is present in 100 g of sample.Hence the initial mass of the sample should be (100X30)/80=37.5 g