The half-life for the reaction $N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$ 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?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

1.5 L

3.36 L

1.05 L

0.07 L

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

$n = n o . o f h a l f - l i f e$

$N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$

Moles of $N_{2} O_{5} l e f t = 0.1 \times {(\frac{1}{2})}^{n} = \frac{0.1}{16}$
Moles of $N_{2} O_{5}$ changed to product $= (0.1 - \frac{0.1}{16}) = \frac{1.5}{16} m o l$

Moles of $O_{2}$ formed $= \frac{1.5}{16} \times \frac{1}{2} = \frac{1.5}{32}$
Volume of oxygen = $1.5 / 32 \times 22.4 = 1.052$