The half-life for the reaction `N_(2)O_(5) to 2 NO_(2) + (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

Options

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

Answer

C

Explanation

Moles of `N_(2)O_(5)=(10.8)/(108)=0.1` and `n=(9.6)/(2.4)=4`
Here `n=` numbers of half life
`N_(2)O_(5) to 2NO_(2)+(1)/(2)O_(2)`
`:.N_(t)=0.1xx((1)/(2))^(n)`
Moles of `N_(2)O_(5)` left `=(0.1)/(16)`
Moles of `N_(2)O_(5)` changed to product
`=0.1-(0.1)/(16)=(1.5)/(16)mol`
Moles of `O_(2)` formed `=(1.5)/(16)xx(1)/(2)=(1.5)/(32)`
Volume of oxygen `=(1.5)/(32)xx22.4=1.05L`.