Correct Answer - C
`NaCO_(3) +2HCI rarr 2NaCI +H_(2)O +CO_(2)`
`NaHCO_(3)+HCI rarr NaCI +H_(2)O +CO_(2)`
If the mixture contains 1 mol of each. `HCI` required is `(2+1) = 3` mol
Molar mass of `Na_(2)CO_(3) = 106 g mol^(-1)`
Molar mass of `NaHCO_(3) = 84 g mol^(-1)`
Thus, `190g (106 g +84g)` of mixture require `HCI = 3` mol
1g of mixture require `HCI =(3)/(190)` mol
`:.` Vol of `0.M HCI` which contains `(3//190)` mol
`= (1000 xx 3)/(190 xx 0.1) = 157.89mL`
