Question

One mole of a mixture of CO and ${CO}_2$ requires exactly 20 g of NaOH in solution for complete conversion of all the ${CO}_2$ into ${Na}_2{CO}_3$. How much NaOH would it require for conversion into ${Na}_2{CO}_3$, if the mixture (one mole) is completely oxidized to ${CO}_2$?

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

Answer 1

The correct option is A 60 g

let $x$ and $y$ mass of $CO$ and $C{O}_2$ be there

$x+y=1$

$2NaOH+C{O}^2\to N{a}_{2}C{O}_3+H_{2}O$

$n$ of $NaOH=20/40=0.5$

$\therefore 0.5/2$ mol $C{O}_2$ must be there

$\Rightarrow 0.25$ mol $C{O}_2$

$\therefore 0.75$ mol $CO$

If $CO$ is oxidised to $C{O}_2$

$NaOH$ required $=2\times 0.75\times 40$

$=60g$