One mole of a mixture of CO and `CO_(2)` requires exactly 20g of NaOH in solution for complete conversion of all the `CO_(2)` into `Na_(2)CO_(3)` if the mixture (one mole is completely oxidised to `CO_(2)`

Options

A. 60g
B. 80g
C. 40g
D. 20g

Answer

B

Explanation

meq. Of `CO_(2)` in mixture `=(20)/(40) xx 1000 = 500`
`:.` mole of `CO_(2)` in mixture
`=(500)/(2 xx 1000) = (1)/(4)` (eq. wt of `CO_(2) = M//2)`
`:.` Mole of CO in mixture `=(3)/(4)`
If this CO is completely oxidised to `CO_(2)` then mole of `CO_(2)` formed `=(3)/(4)`
`:.` Total mole of `CO_(2) = (1)/(4)+(3)/(4) =1`
`:.` Mole of NaOH required =`2 xx` mole of `CO_(2) = 2 xx 1 = 2`
`:.` wt of `NaOH` required `= 2xx 49 = 80g`