How many `mL` of a `0.1M HCl` are required to react completely with `1 g` mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of two?

Answer

`157.8mL`

Explanation

Let moles of `Na_(2)CO_(2)` and `NaHCO_(3)` in `L` of mixture `=a`
For reaction `:('n'` factor for `Na_(2)CO_(3)=2` and for `NaHCO_(3)=1)N=Mxx'n'` factor
`N_(1)V_(1)` of `Na_(2)CO_(3)+N_(2)V_(2)` of `NaHCO_(3)=N_(1)xxV_(1)` of `HCl ('n'` factor `=1)`
`axx2xx1000+axx1xx1000=0.1xxV`
`:. 3a=10^(-4)xxV` ......(i)
`Mw` fo `Na_(2)CO_(3)=106` and `Mw` of `NaHCO_(3)=84g. mol^(-1)`
Since of misture `=1g`
`:. ` Weight of `Na_(2)CO_(3)+` Weight of `NaHCO_(3)=1g`
or `a xx 106+a xx 84=1(W=Mw xx "moles")`
`:.a=5.26xx10^(-3)`.......(ii)
From Eqs. (i) and (ii),
`3xx5.26xx10^(-3)=10^(-4)V`
or `V=157.8mL`