1. 84 g of a mixture of CaCO_(3) nad Mg CO_(3) was heated to a constant weight. The constant weightof theresiduewas found to be 0.96g. Calculate thepercentage compositionofthemixture. ( Ca = 40 , Mg = 24 , c = 1 , o = 16)

1. 84 g of a mixture of CaCO_(3) nad Mg CO_(3) was heated to a constan

1.	1. 84 g of a mixture of CaCO_(3) nad Mg CO_(3) was heated to a constant weight. The constant weightof theresiduewas found to be 0.96g. Calculate thepercentage compositionofthemixture. ( Ca = 40 , Mg = 24 , c = 1 , o = 16)
Answer» Solution :On heating `caCO_(3) and MgCO_(3)` , one of the products , `CO_(2)` , escapes out . We have, `{:(CaCO_(3)+MgCO_(3)""to ""CaO+MgO+CO_(2)uarr),("x g(-84 - x) gy g (0.96 - y ) g "),("(say)(say)"):}` APPLYING POAC for Ca atoms, moles of Ca atoms in `CaCO_(3)` = moles of Ca atomsin CaO `1 xx ` moles of `CaCO_(3) = 1 xx ` molesof CaO `(x)/(100) = (y)/( 56)""[{:(CaCO_(3)=100),(" "CaO = 56):}]. . . (i)` Again applying POAC for Mg atoms, moles of `MgCO_(3) ` = molesof Mg in MgO `1 xx` molesof `MgCO_(3)= 1 xx ` moles of MgO `(1 . 84 - x)/( 84) = (0.96 - y)/( 40 ) [{:(MgCO_(3)=84),(" MgO = 40 "):}]"". . . (ii)` From eqns . (i) and (ii), we GET x = 1 g, y = 0. 84 g `% of Ca CO_(3)= (1)/( 1 . 84) xx 100 = 54.34%` and `% of MgCO_(3) = 45.66%` Second Method Applying POAC for C atoms.