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. 60 g
B. 80 g
C. 200 G
D. 263 G

Answer

B

Explanation

`"meq of "CO_(2)" in mixture "=(20)(40)xx1000=500`
`therefore " mole of "CO_(2)" in mixture"`
`=(500)/(2xx1000)=(1)/(4) ("Eq. wt of " CO_(2)=M//2)`
`therefore" Mole of CO in mixture "=(3)/(4)`
If this `CO` is completely oxidised to `CO_(2)` then mole of `CO_(2)` formed `=(3)/(4)`
`therefore " Total mole of "CO_(2)=(1)/(4)+(3)/(4)=1`
`therefore " Mole of NaOH required "=2xx" mol of "CO_(2)=2xx1=2`
`therefore" wt. of NaOH required "=2xx40=80 g`