1.

1.84 g of a mixture of CaCO_(3) and MgCO_(3) is strongly heated till no further loss of mass takes place. The residue weighs 0.96 g. Calculate the percentage composition of the mixture.

Answer»

Solution :Suppose the mass of `CaCO_(3)` in the mixture = x g.
`THEREFORE` Mass of `MgCO_(3)` in the mixture will be = (1.84 - x) g.
Step 1. To calculate the mass of CAO RESIDUE from x g calcium carbonate.
`underset(100 g)(CaCO_(3))overset(Delta)rarr underset(56 g)(CaO)+CO_(2)`
Thus, 100 g of `CaCO_(3)` upon decomposition give a residue of CaO = 56 g
`therefore` x g of `CaCO_(3)` will give residue of `CaO=(56xx x)/(100)=0.56xx x g`
Thus, the mass of CaO residue formed = 0.56 x g
Step 2. To calculate the mass of MGO residue from (1.84 - x) of magnesium carbonate.
`underset(84 g)(MgCO_(3))overset(Delta)rarr underset(40 g)(MgO+CO_(2)`
84 g of `MgCO_(3)` upon decomposition yield residue of MgO = 40 g
`therefore (1.84 - x)g` of `MgCO_(3)` will yield residue of `MgO=(40xx(1.84-x))/(84)g`
Thus, the mass of MgO residue formed `=(40)/(84)(1.84-x)g`.
Step 3. To calculate the masses of `CaCO_(3)` and `MgCO_(3)` in the mixture.
Total mass of CaO and MgO residue from Step 1 and Step 2
`=0.56 x + (40)/(84)(1.84 - x)=0.96 g` (Given)
or `0.56 x xx 84+40xx1.84-40x=0.96xx84`or `47.04x -40x -40x=80.64-73.60`
or `7.04x=7.04`or x = 1
Thus, the mass of `CaCO_(3)` in the mixture = 1.00 g
and the mass of `MgCO_(3)` in the mixture `= 1.84-1.00 = 0.84 g`.
% of `CaCO_(3)` in the mixture `=(1)/(1.84)xx100=54.35`
% of `MgCO_(3)` in the mixture = 100 - 54.35 = 45.65.


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