A mixture of `NaHCO_(3) and Na_(2)CO_(3)`, weighed 1.0235. The dissolved mixture was reached with excess of `Ba(OH)_(2)` to form 2.1028g `BaCO_(3)`, by the following reactions:
`Na_(2)CO_(3)+Ba(OH)_(2)toBaCO_(3)+2NaOH`
`NaHCO_(3)+Ba(OH)_(2) to BaCO_(3)+NaOH+H_(2)O`
What was the percentage of `NaHCO_(3)` in the orginal mixture?

Explanation

Let x g of `NaHCO_(3)` be present in the mixture?
Mass of `Na_(2)CO_(3)` in the mixture=(1.025-x)g`
Number of moles of `NaHCO_(3)=(x)/(84)`
Number of moles of `Na_(2)CO_(3)`=(1.0235-x)/(106)`
Number of moles of `BaCO_(3)`=Number of moles of `NaHCO_(3)+"Number of moles of" Na_(2)CO_(3)`
`(2.1028)/(197)=(x)/(84)+(1.0235-x)/(106)`
x=0.4122
`"Amount of " NaHCO_(3)=0.4122g`
`"Percentage of "NaHCO_(3)=(0.4122)/(1.0235)xx100=40.27`