A sample contains a mixtrure of `NaHCO_(3)` and `Na_(2)CO_(3)`.
`HCl` is added to `15.0 g`.of the sample, yielding `11.0 g` of `NaCl`. What percent of the sample is `Na_(2)CO_(3)`?
`[{:("Reaction are"),(Na_(2)CO_(3) + 2HCl rarr 2NaCl + CO_(2) + H_(2) O),(NaHCO_(2) + HCl rarr NaCL + CO_(2) + H_(2) O):}]`
`Mw "of" NaCl = 58.5, Mw "of" NaHCO_(3) = 84, Mw "of" Na_(2)CO_(3) = 106 g mol^(-1)`

Explanation

Let `x` of `Na_(2)CO_(3)`. Then, weight of `NaHCO_(3) = (15 - x) g`
Moles of `NaCl` produced `= (11.0 g)/(58.5 g) = 0.188"mol"`
The `NaCl` is produced by the reaction of `((x)/(106)) "mol"`
of `Na_(2) CO_(3)` and `((15 - x))/(84) "mol of" NaHCO_(3)`. Each mol of `Na_(2)CO_(3)` produces 2 mol of `NaCl`
`:. (2x)/(106) + (15 - x)/(84) = 0.188`
Solve `x : = 13.5 g Na_(2) CO_(3)`,
`NaHCO_(3) = (15 - 1.35) = 13.6 g`
`% Na_(2)CO_(3) = (1.35)/(15) xx 100 = 9.0% Na_(2) CO_(3)`