1. 84 g of a mixture of `CaCO_(3) nad Mg CO_(3)` was heated to a constant weight . The constant weight of the residue was found to be 0.96 g. Calculate the percentage composition of the mixture. ( Ca = 40 , Mg = 24 , c = 1 , o = 16)

Explanation

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) g y g (0.96 - y ) g "),(" (say) (say)"):}`
Applying POAC for Ca atoms,
moles of Ca atoms in `CaCO_(3)` = moles of Ca atoms in CaO
`1 xx ` moles of `CaCO_(3) = 1 xx ` moles of CaO
`(x)/(100) = (y)/( 56)" "[{:(CaCO_(3)=100),(" "CaO = 56):}] . . . (i)`
Again applying POAC for Mg atoms,
moles of `MgCO_(3) ` = moles of Mg in MgO
`1 xx` moles of `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.