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)`

Answer

5

Explanation

Use `M=("% by weight "xx10xxd)/(Mw_(2))`
Mol. Wt. of `C_(2)H_(5)OH=46 gm`
`M_(1)V_(1)=M_(2)V_(2)`
`(90xx10xx0.8)/46 xxV=(10xx10xx0.9)/46 xx40`
`V=(10xx0.9xx40)/(90xx0.8)=5 mL`.