1 gram of mixture of carbonates of calcium and magnesium gave 240 cm^3 of carbon dioxide at S.T.P. Calculate the percentage composition of the mixture.

Asked by Manoj | 8th Jul, 2013, 08:33: PM

Expert Answer:

1. Let the mass of CaCO3 in the mixture be X g.
   Let the mass of MgCO3 in the mixture be (1 - X )g.
2. calculate volume of CO2 from  CaCO3 -
    CaCO3 ?  CaO  +  CO
      100 g                22400 mL
   100 g of CaCO3 forms 22400 mL of CO2
    x g of CaCOforms     22400 x  / 100   =  224 x mL.
 
3. calculate volume of CO2 from  MgCO3 -
    MgCO3 ?  MgO  +  CO
      84 g                 22400 mL
   84 g of MgCOforms 22400 mL of CO2
    (1 - x) g of MgCOforms     22400 (1 -x ) / 84   =  266.67 ( 1-x) mL.
4. To calculate % 
    Total  CO2 evolved  = 224 x  mL  + 266.67 ( 1-x) mL.
   Actually 240 mL  COevolved ;
Hence    224 x  mL  + 266.67 ( 1-x) mL.  = 240 mL 
224x  + 266.67 + 266.67x  = 240
or x = 26.67  / 42.67     
     = 0.6250 
 
thus  % CaCOin the mixture  = 0.6250  x 100
                                         = 62.50 %
% MgCOin the mixture  = 100 - 62.50
                                  = 37.50 %

Answered by  | 13th Jul, 2013, 12:55: AM

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