The half-life for the reaction  $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$ is 2.4 h at STP. Starting with 10.8g of  $N_2{O}_5$,  how much oxygen will be obtained after a period of 9.6h?

Given  $t=9.6h,n=\frac{9.6}{2.4}=4$

$n=no.ofhalf-life$

$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$

Moles of  $N_2{O}_{5}left=0.1\times {\left(\frac{1}{2}\right)}^n=\frac{0.1}{16}$
Moles of $N_2{O}_5$ changed to product $=\left(0.1-\frac{0.1}{16}\right)=\frac{1.5}{16}mol$
 
Moles of $O_2$  formed  $=\frac{1.5}{16}\times \frac{1}{2}=\frac{1.5}{32}$
Volume of oxygen =  $1.5/32\times 22.4=1.052$