Question

After 11.2 g of carbon reacts with oxygen originally occupying 21.2 lit at ${18}^{o}C$ and 750 torr, the cooled gases are passed thorough 3 litre of 2.5 M $NaOH$ solution. Determine the concentration of $NaOH$ remaining in solution which is not converted to ${Na}_{2}C{O}_3$.
Note : $CO$ does not react with $NaOH$ under these conditions.

Answer 1

Mole of carbon $=\frac{11.2}{12}=0.933mole$

Mole of $O_2=\frac{PV}{RT}=\frac{\left(750/760\right)\times 21.2}{0.0821\times (273+18)}=0.88moles$

$\begin{array}{cccccc}Step1& C& +& O_2& \to & CO\\ t=0& 0.93& & 0.88& & 0.88\\ t& 0.05& -& & & 0.88\end{array}$

$\begin{array}{cccccc}Step2& CO& +& O_{2_{\left(temaaining\right)}}& \to & {CO}_2\\ t=0& & & O& & O\end{array}$

$\because$ As $NOC{O}_2$ is formal

$\therefore$ No change in comentration of $NaOH$