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A `2.0 g` sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of `CO_(2)` ceases. The volume of `CO_(2)` at `750mm Hg` pressure and at `298 K` is measured to be `123.9 mL`. A `1.5 g` of the same sample requires `150 mL` of `(M//10) HCl` for complete neutralisation. Calculate the percentage composition of the components of the mixture. |
Answer» Only `NaHCO_(3)` decoposes to give `CO_(2)`. `2NaHCO_(3)overset(Delta)toNa_(2)CO_(3)+CO_(2)+H_(2)O` `PV=nRT` or `(750)/(760)xx(123.9)/(1000)=nxx0.0821xx298` or `n=4.997xx10^(-3)approx5xx10^(-3) mol CO_(2)` Moles of `NaHCO_(3)=5xx10^(-3)xx2=0.01` Mass of `NaHCO_(3)=0.01xx84=0.84g` Percentage of `NaHCO_(3)=(0.84)/(2)xx10=42%` Millimoles of `HCl=150xx(1)/(10)=15` HCl reacts with `Na_(2)CO_(3)` and `NaHCO_(3)`. `1.5 g` mixture has `(0.01)/(2)xx1.5=0.0075` mol `NaHCO_(3)` `=7.5 m" mol of "NaHCO_(3)` `NaHCO_(3)+"HCl"toNaCl+CO_(2)+H_(2)O` Millimoles of HCl neutralised by `Na_(2)CO_(3)=15-7.5=7.5` `Na_(2)CO_(3)+2"HCl"to2NaCl+CO_(2)+H_(2)O` Millimoles of `Na_(2)CO_(3)=(7.5)/(2)=3.75` Mass of `Na_(2)CO_(3)=3.75` Mass of `Na_(2)CO_(3)=3.75xx106xx10^(-3)=0.3975g` Percentage of `Na_(2)CO_(3)=(0.3975)/(1.5)xx100=26.5%` Percentage of `Na_(2)SO_(4)=100-(42+26.5)=31.5%` |
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