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?

Explanation

Step I. Calculation of mass of constituents in the mixture
Mass of mixture = 1.0 g
Let the mass of `Na_(2)CO_(3) = xg`
Mass of `NaHCO_(3)=(1-x)g`
Moles of `Na_(2)CO_(3)=("Mass of "Na_(2)CO_(3))/("Molar mass")=((xg))/(("106 g mol"^(-1)))=(x)/(106)` mol
Moles of `NaHCO_(3)=("Mass of "NaHCO_(3))/("Molar mass")=((1-x)g)/(("84 g mol"^(-1)))=(1-x)/(84)`mol
According to available data :
Moles of `Na_(2)CO_(3)` = Moles of `NaHCO_(3)`
`:. ((x)/(106)"mol")=((1-x)/(84)"mol")`
or `84x=106-106xorx=(106)/(190)=0.558g`
`:.` Mass of `Na_(2)CO_(3)` in the mixture = 0.558 g
Mass of `NaHCO_(3)` in the mixture `= (1 - 0.558) =0.442 g`
Step II. Calculation of total mass of `HCl` required
`underset(106g)(Na_(2)CO_(3)(s))+underset(73g)(2HCl)(aq)rarr2NaCl(aq)+H_(2)O(l)+CO_(2)(g)`
`underset(84g)(NaHCO_(3))(s)+underset(36.5g)(HCl(aq))rarrNaCl(aq)+H_(2)O(l)+CO_(2)(g)`
Now, 106 g of `Na_(2)CO_(3)` require HCl = 73 g
0.558 g of `Na_(2)CO_(3)` require `HCl=((73g)xx(0.558g))/((106g))=0.384g`
Similarly, 84 g of `NaHCO_(3)` require HCl = 36.5
0.442 g of `NaHCO_(3)` require HCl `= ((36.5g)xx(0.442g))/((84g))=0.192g`
`:.` Total mass of HCl required `= (0.384 + 0.192) = 0.576 g`
Step III. Calculation of volume of `HCl` required
Mass of HCl required = 0.576 g
Molarity of HCl solution = 0.1 M
Molarity of solution (M) `= ("Mass of HCl/Molar mass of HCl")/("Volume of HCl solution (V)")`
`("0.1 mol L"^(-1))=((0.576g)//(36.5g//mol))/(V)`
`= V = ((0.576g))/(("36.5 gmol"^(-1))xx("0.1 mol L"^(-1)))=0.1578L =157.8mL`.