MgO+CO2(g) Mole ratio 1: 1: 1 Mass ratio (ing) 84: 40: 44 Given values 10g 4g 0.1mol (=4.4g){∵n=‌
w
m
} (i) ∵84g. of MgCO3 give MgO=40g ∴10g of MgCO3 give MgO=‌
40×10
84
=4.76g Thus, MgO obtained is less by 4.76−4.0=0.76g (ii) ∵84g of MgCO3 give CO2=44g ∴10g of MgCO3 give CO2=‌
44×10
84
=523g Thus CO2 obtained is less by 5⋅23−4⋅4=0⋅84g Total short =0⋅76+0⋅84=1⋅59=1.60 ∵10g. of MgCO3 give less products by =1.60g ∴100g of MgCO3 give less products by =1.600g Hence, MgCO3=100−16=84%