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 1. Calculating the number of grams of the components in the mixture:
Let x g be the mass of `Na_(2)CO_(3)` in the mixture, so that `(1-x) g` becomes the mass of `NaHCO_(3)` in 1g mixture.
Formula mass `(Na_(2)CO_(3))=106 u//"formula"` unit
Molar mass `(Na_(2)CO_(3))=106 g mol^(-1)`
Formula mass `(NaHCO_(3))=84 u//"formula"` unit
Molar mass `(NaHCO_(3))=84 g mol^(-1)`
Thus, `n_(Na_(2)CO_(3))=(xg)/(106 g mol^(-1))`
`n_(NaHCO_(3))=((1-x)g)/(84 g mol^(-1))`
Since mixture contains equimolar amounts of both, we have
`n_(Na_(2)CO_(3))=n_(NAHCO_(3))`
`(xg)/(106 g mol^(-1))=((1-x)g)/(84 g mol^(-1))`
`106(1-x)=84 x`
`106-106 x=84 x`
`106=84x+106x=190x`
`or x=106/190g=0.558 g`
and `1-x=1-0.558=0.4425 g `
Step 2. Finding moles of the components in the mixture:
`n_(Na_(2)CO_(3))=(0.558 g)/(106 g mol^(-1))=0.00526 mol`
`n_(NaHCO_(3))=(0.442 g)/(84 g mol^(-1))`
`=0.00526`
Note that both should have equal moles
Step 3. Finding the moles of HCl through the balanced chemical equation
`Na_2CO_3(s)+2HCl(aq.)to2NaCl(aq.)+H_2O(l)+CO_2(g)`
According to the balanced equation, the number of moles of HCl conumed is twice the number of moles of `Na_(2)CO_(3)` consumed:
Moles of `HCl=0.00526 mol Na_(2)CO_(3)xx(2 mol HCl)/(1 mol Na_(2)CO_(3)`
`=0.01052 mol HCl`
Next equation is
`NaHCO_(3)(s)rarr HCl(aq.) rarr NaCl(aq.)+H_(2)O(1)+CO_(2)(g)`
According to the balanced equation, the number of moles of HCl consumed is the same as that of NaOH consumed:
Moles of `HCl=0.00526 mol NaHCO_(3)xx(1 mol HCl)/(1 mol NaHCO_(3))`
`=0.00526 mol HCl`
Thus, total moles of HCi needed for complete reaction with the given mixture:
`n_(HCl)=(0.01052+0.00526) mol HCl`
`=0.01578 mol HCl`
Step 4. Dividing the number of moles of HCl by the molarity of HCl gives the volume of HCl:
Volume of `HCl=n_(HCl)/M=(0.01578 mol HCl)/(0.1 mol HCl//L "soln")`
`=0.01578 L "soln"`
`=(0.01578 L "soln")((1000 mL "soln")/(1 L "soln"))`
`=15.78 mL "soln"`