How many ml of 0.1 HCl is required to react completely with 1.0g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equi-molar amounts of both ?

Explanation

Let `a` moles `Na_(2)CO_(3)` and `a` moles of `NaHCO_(3)` be present in `1 g` mixture.
`:. axx106+axx84=1`
or `a=5.26xx10^(-3)`
Now for reaction:
`"Meq of" HCl="Meq.of"Na_(2)CO_(3)+"Meq of"NaHCO_(3)`
`0.1xx1xxV=2xx5.26xx10^(-3)xx1000+1xx5.26xx10^(-3)xx1000`
`V=157.8mL`