Like always, start with a balanced chemical equation for this decomposition reaction
#"CaCO"_text(3(s]) stackrel(color(red)(Delta)color(white)(aa))(->) "CaO"_text((s]) + "CO"_text(2(g])# #uarr#
Notice that you have
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
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
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
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
If the volume of CO2 ie 6.72L is in STP, then calculation may be made as follows
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