Question

1 g of a mixture of carbonates of calcium and magnesium gave 240cm^3 of CO2 at STP. Calculate percentage composition of mixture​

Answer 1

Answer:

CaCO3 =  0.625 mg

  MgCO3 = 0.375 mg

Explanation:

CaCO3--> CaO  + CO2

100 gm CaCO3 gives 1 mole of CO2

Again, MgCO3--> MgO  + CO2

84 gm MgCO3 gives 1 mole of CO2

Say Mixture contains x gm CaCO3

So  x gm CaCO3 gives= x/100 mole CO2

(1-x ) gm MgCO3 gives = (1-x)/84 mole of CO2

Again 240 c.c CO2 at STP = 240/22400 mole of CO2

so equation is,

x/100 +  (1-x)/84  =  240/22400

x=5/8 gm = 0.625 mg

so CaCO3 =  0.625 mg

  MgCO3 = 0.375 mg