Question

$200ml$ of $C_2{H}_4$ is burnt in just sufficient air (containing $20\%$ oxygen). Calculate the resultant mixture composition at ${100}^{\circ }C$.

Answer 1

Since air contains mainly $N_2$ and $O_2$.

$\%O_2=20\%$

$\%N_2=100-20=80\%$

Now,

$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$

$V$ $3V$ $2V$ $2V$

Used $:200ml600ml400ml400ml$

Left $:0ml0ml$

$\therefore$ Now,

Total volume of air $=xml$

Again,

$O_2\to 20\%$ of $x=600ml$

$x=\frac{600\times 100}{20}ml$

$=3000ml$

$\therefore$ Volume of $N_2$ in mixture $=80\%$ of $3000ml$

$=2400ml$

$\therefore$ $\left(i\right)V_{N_2}=2400ml$

$\left(ii\right)V_{C{O}_2}=400ml$

$\left(iii\right)V_{H_{2}O}=400ml$