Question

$11.2$ g carbon reacts completely with $19.63$ litres of $O_2$ at NTP. The cooled gases are passed through $2$ litres of $2.5$ $N$ $NaOH$ solution. The concentration of remaining ${Na}_{2}C{O}_3$ in solution is (as nearest integer) ($CO$ does not react with $NaOH$ under these conditions) :

Answer 1

The reactions are as follows:

$C+\frac{1}{2}{O}_2\to CO$
$a$ mol
$C+O_2\to C{O}_2$
$b$ mol

Volume of $O_2$ that reacts with $C$ to give $CO+$ volume of $O_2$ that reacts with $C$ to give $C{O}_2$ $=19.63$ litres

$\frac{a}{2}\times 22.4+b\times 22.4=19.63...\left(i\right)$

$⟹$ $11.2a+22.4b=19.63$

Also, moles of $C=a+b$

$\therefore$ Mass of $C$ used $=12(a+b)$
or $12a+12b=11.2...\left(ii\right)$

Solving Eqs. $\left(i\right)$ and $\left(ii\right)$, we get $b=0.82$
$a=0.11$

$\therefore$ Moles of $C{O}_2$ formed $=$ Moles of $C$ used for $C{O}_2$ $=0.82$

Meq. of $NaOH$ solution in which $C{O}_2$ is passed =$2000\times 2.5=5000$

Meq. of $C{O}_2$ passed $=0.82\times 1000\times 2=1640$

$\therefore$ Meq. of $NaOH$ left $=5000-1640=3360$

$\therefore N_{NaOH}$ left $=\frac{3360}{2000}=1.68N$

Meq. of ${Na}_{2}C{O}_3$ formed $=$ Meq. of $C{O}_2$ passed $=1640$

$\therefore N_{{Na}_{2}C{O}_3}=\frac{1640}{2000}=0.82N$

So, the nearest integer is $1$.