One gram of a mixture of `NaHC_2O_4` and `Na_2C_2O_4` dissolved in water consumes in titration 50.28 " mL of " 0.1 N NaOH with phenolphthalein as indicator. If 1 g of the mixture is heated to constant weight, what would be the weight of `Na_2CO_3` produced?

Explanation

The acid oxalate consumes 50.28 " mL of " 0.1 N NaOH
50.28 " mL of " 0.1 N NaOH
`=(50.28)/(1000)xx0.1Eq=5.028xx10^(-3)Eq`
This reacts with the same quantity i.e., `5.028xx10^(-3)mol`
of `NaHC_2O_4=(5.028xx10^(-3)xx112)g=0.563g`
`1-0.563=0.437 ` g is `Na_2C_2O_4`
Since `2NaHC_2O_4` gives `Na_2CO_3,0.563` g would give
`(0.563)/(224)xx106g=0.2664` g of `Na_2CO_3`
`underset(134g)(Na_2C_2O_4)tounderset(106g)(Na_2CO_3)+CO`
0.437 g would yield `(0.437)/(134)xx106=0.3457g of Na_2CO_3`
Total wewight of `Na_2CO_3=0.2664+0.3457`
`=0.6121 g`