Question

caco3 - CaO + CO2
(100gm) (56gm) (44gm)
In this reaction, How much CO2 is liberated if
10 gm of calcium carbonate is heated.​

Answer 1

Answer:

The reaction that is happening is this:

CaCO3 → CaO + CO2

So firstly we need to find the molar mass of CaCO3:

Mr(CaCO3) = Ar(Ca) + Ar(C) + 3*Ar(O)

Mr(CaCO3) = 40.078 + 12.0107 + 3*15.9994

Mr(CaCO3) = 52.0887 + 47.9982

Mr(CaCO3) = 100.0869

M(CaCO3) = 100.0869 g/mol

After that we will find the moles of the calcium carbonate:

n(CaCO3) = m(CaCO3)/ M(CaCO3)

n(CaCO3) = 100g/ 100.0869 g/mol

n(CaCO3) = 0.9991 mol

Now we need to find the molar attitude between the calcium carbonate and the carbon dioxide:

n(CaCO3) : n(CO2) = 1 : 1

Now from this we can find the moles of the carbon dioxide:

n(CO2) = n(CaCO3)

n(CO2) = 0.9991 mol

Now we need to find the molar mass of the carbon dioxide:

Mr(CO2) = Ar(C) + 2*Ar(O)

Mr(CO2) = 12.0107 + 2*15.9994

Mr(CO2) = 12.0107 + 31.9988

Mr(CO2) = 44.0095

M(CO2) = 44.0095 g/mol

Finally we can calculate the mass of the produced CO2:

m(CO2) = n(CO2) * M(CO2)

m(CO2) = 0.9991 mol * 44.0095 g/mol

m(CO2) = 43.9698 g

m(CO2) = 44 g

So in conclusion the mass of the formed CO2 is about 44 grams.

Hope, that that helps! :)