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?

A mixture of `NaHCO_(3) and Na_(2)CO_(3)`, weighed 1.0235. The dissolv

1.	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?
Answer» 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`

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