Volume of `0.1M HCI` required to react completely with 1g equimolar mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` is

Explanation

Let the amount of `Na_(2)CO_(3)` be xin the mixture .
Amount of `NaHCO_(3) = 1 -x`
Moles of `Na_(2)CO_(3) = (x)/(106)` (Molar mass of `Na_(2)CO_(3) = 106`)
Moles of Na`HCO_(3) = (1-x)/(84)`
(Molar mass of `NaCHO_(3) = 84`)
Since number of moles of both are equal.
` (x)/(106) = (1-x)/(84)`
` 84 x = 106 - 106 x`
` 190 x = 106 therefore x = 0.558`
`therefore ` Moles of `Na_(2)CO_(3) = (0.558)/(106) = 0.00526`
Moles of `NaCHO_(3) = (1-0.558)/(84)`
`= 0.00526` ( Moles are equal)
Now , HCl react with `Na_(2) CO_(3)` as :
`Na_(2)CO_(3) + 2HCl to 2NaCl + H_(2)O + CO_(2)`
`NaHCO_(3) + HCl to NaCl + H_(2)O + CO_(2)`
According to the reactions 1, mol of `NaCO_(3)` will react with 2 mol of HCl and therefore m 0.00526 mol of `Na_(2)CO_(3)` will react with `2 xx 0.00526`
mol of HCI.Similary .
1 mol of `NaHCO_(3)` will required to react with 0.00526 mol of HCl. Total moles of HCl required to rect with mixture of `NaHCO_(3)` and `Na_(2)CO_(3)` .
` = 2 xx 0.00526 + 0.00526`
` = 0.01578` mol.
To calculated volume of 0.1 M HCl correcponding to 0.001578 mol, we have .
0.1 mol of 0.1 M HCl is present in 1000 mL.
0.01578 mol of 0.1 M HCl is prsent in `(1000)/(0.1) xx 0.01578 `
`= 157.8mL`