Question

$50mL$ of $CO$ is mixed with $20mL$ of oxygen and sparked. After the reaction, the mixture is treated with an aqueous $KOH$ solution. Choose the correct option:

• A. the volume of $CO$ that reacts $=30mL$
• B. volume of ${CO}_2$ formed $=50mL$
• C. volume of $CO$ that remains after treatment with $KOH=10mL$
• D. the volume of the $CO$ that remains after treatement with $KOH=20mL$

Answer 1

The correct option is C volume of $CO$ that remains after treatment with $KOH=10mL$

Given,

$50ml$ $CO$ is mixed with $20ml$ Oxygen and sparked.

The reaction involved is

$2CO+O_2⟶2C{O}_2$

ratio is $2:1:2$

$\therefore$ If $20ml$ oxygen is used, then it consumes $40ml$ $CO$ to produce $40ml$ $C{O}_2⟶\left(1\right)$

as Oxygen here is limiting reagent.

Now further this mixture is treated with aqueous $KOH$ solution

The reaction involved here is

$2KOH+C{O}_2⟶K_{2}C{O}_3+H_{2}O$

Ratio $2$ : $1$ : $1$ : $1$

$\therefore 40ml$ $C{O}_2$ will react with $80ml$ $KOH$ and produce $40ml$ $K_{2}C{O}_3$ $⟶\left(2\right)$

From (1) & (2),

Finally $(50ml-40ml)$ $CO$ is remained i.e; $10ml$ $CO$ remained along with $40ml$ $K_{2}C{O}_3$ & $40ml$ $H_{2}O$.