CaCO3(s)→ΔCaO(s)+CO2(g)
MgCO3(s)→ΔMgO(s)+CO2(g)
Let the weight of CaCO3 be xgm
∴ weight of MgCO3=(2.21−x)gm
Moles of CaCO3 decomposed = moles of CaO formed
100x= moles of CaO formed
∴ weight of CaO formed =100x×56
Moles of MgCO3 decomposed = moles of MgO formed
84(2.21−x)= moles of MgO formed
∴ weight of MgO formed =842.21−x×40 ⇒842.21−x×40+100x×56=1.152
∴x=1.187g= weight of CaCO3 and weight of MgCO3=1.023g