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3.75 g of a mixture of CaCO3 and MgCO3 i...

3.75 g of a mixture of `CaCO_3` and `MgCO_3` is dissolved in 1 L of 0.1 M HCl to liberate 0.04 " mol of "`CO_2`. Calculate
(a). The percentage of each compound in the mixture.
(b). The amound of acid used.
(c). The amound of acid left after the reaction.

Text Solution

Verified by Experts

(a). Let the weight of `CaCO_(3)` and `MgCO_(3)` be x and y
respectively
`thereforex+y=3.75` ....(i)
`CaCO_(3)+2"HCl" toCaCl_(2)+H_(2)O+CO_(2)`
`MgCO_(3)+2H"Cl"toMgCl_(2)+H_(2)O+CO_(2)`
`[Mw(CaCO_(3))=100g` and Mw `(MgCO_(3))=84g]`
" mol of "`CaCO_(3)+` " mol of "`MgCO_(3)=` " mol of "`CO_(2)` formed
`therefore(x)/(100)+(y)/(84)=0.04`
Solving equation (i) and (ii) we get
`x=2.4375g`
`y=3.75-2.4375=1.3125g`
% of `CaCO_(3)=(2.4375xx100)/(3.75)=65%`
% of `MgCO_(3)=100-65=35%`
(b). HCl used `=2xx" mol of "CO_2=2xx0.04=0.08mol`
(c). HCl left =" mol of "HCl added `-`" mol of "HCl used
`=1Lxx0.1-0.08=0.02mol`
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