KEAM 2017 Physics and Chemistry Paper

since, MgCO3

∆

⟶

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%