2007 ce chem mc

First calculate the % by mass of C in CaCO3:
12.0 / 100.1 x 100% = 12.0%
As there is an increase in volume of CO2 collected, the no. of moles of CO2 is greater. Thus the impurity must be a metallic carbonate with % by mass of C greater than 12.0%.
For MgCO3, % by mass of C: 12.0 / 84.3 x 100% = 14.2%
For ZnCO3, % by mass of C: 12.0 / 125.4 x 100% = 9.57%
For FeCO3, % by mass of C: 12.0 / 115.8 x 100% = 10.4%
For CuCO3, % by mass of C: 12.0 / 123.5 x 100% = 9.72%
Hence A is the answer.