Question

The half-life for the reaction $N _{2} O _{5} \longrightarrow 2 NO _{2}+\frac{1}{2} O _{2}$ is $2.4\, h$ at STP. Starting with $10.8\, g$ of $N _{2} O _{5}$ how much oxygen will be obtained after a period of $9.6\, h$ ?

• 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.
$N _{2} O _{5} \longrightarrow 2 NO _{2}+\frac{1}{2} O _{2}$
$\therefore N_{t}=0.1 \times\left(\frac{1}{2}\right)^{n}$
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 $=\frac{1.5}{32} \times 22.4$
$=1.05\, L$