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

Options

A. `100mL`
B. `200mL`
C. `157.9mL`
D. `159.7 mL`

Answer

C

Explanation

`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`