Question

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

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

Answer 1

The correct option is C 1.05 L

Given $t=9.6h$, $n=\frac{9.6}{2.4}=4$

Here n = numbers of half-life

$N_2{O}_5\to 2N{O}_2+1/2O_2$ $N_2=0.1\times (1/2{)}^n$

Moles of $N_2{O}_5$ left $=\frac{0.1}{16}$

Moles of $N_2{O}_5$ changed to product $=(0.1-\frac{0.1}{16}$)=$\frac{1.5}{16}$ mol

Moles of $O_2$ formed $=\frac{1.5}{16}\times 1/2=\frac{1.5}{32}$

Volume of oxygen $=1.532\times 22.4=1.05L$