Question

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

• A. 1.5 L
• B. 3.36 L
• C. 1.05 L
• D. 0.07 L

Answer 1

Answer: (C)

Moles of $N_2{O}_5=\frac{10.8}{108}=0.1$ and, $n=\frac{9.6}{2.4}=4$
here $n=$ numbers of half-life.
$N_2{O}_5⟶2N{O}_2+\frac{1}{2}{O}_2$
$\therefore N_t=0.1\times {\left(\frac{1}{2}\right)}^n$
Moles of $N_2{O}_5$ left $=\frac{0.1}{16}$
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 $=\frac{1.5}{32}\times 22.4$
$=1.05L$