Question

How many liters of carbon dioxide will be formed from the decomposition of 14.1g of calcium carbonate (at stp)?

Answer 1

Firstly, according to Ideal Gas Law the volume of 1 mole of a gas at Standard Temperature and Pressure (STP) is 22.4
`dm^(3)`, ie the value of the molar volume (Vm) is 22.4
`dm^(3)`/mol.

`CaCO_(3) = CaO + CO_(2)`

From the reaction, it can be seen that
`CaCO_(3)` and
`CO_(2)` have the following amount of substance relationship:

`n(CaCO_(3)) = n(CO_(2)) = 1:1`

From the relationship we can determinate moles of
`CO_(2)` :

`n(CaCO_(3)) = n(CO_(2)) = n/M= 14.1/100=0.141 moles`

Finally, we can calculate the volume of formed CO2:

V(
`CO_(2)` )=nxVm=0.141x22.4=3.16
`dm^(3)`