For the following reaction : `N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2NO_(2)(g)+O_(3)(g)` `NO_(2)(g)+O_(2)(g)overset(50%)rarrNO(g)+O_(3)(g)` If initial

Question

For the following reaction : `N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2NO_(2)(g)+O_(3)(g)` `NO_(2)(g)+O_(2)(g)overset(50%)rarrNO(g)+O_(3)(g)` If initial

For the following reaction :
`N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2NO_(2)(g)+O_(3)(g)`
`NO_(2)(g)+O_(2)(g)overset(50%)rarrNO(g)+O_(3)(g)`
If initially 20 moles of `N_(2)O_(5)` and 30 moles of `O_(2)` are taken then calculate sum of moles of `O_(2)` and `O_(3)` after the reaction.
A. 16
B. 21
C. 27
D. 30

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2NO_(2)(g)+O_(3)(g)`
`NO_(2)(g)+O_(2)(g)overset(50%)rarrNO(g)+O_(3)(g)`
For `1^(st)` reaction: `N_(2)O_(5)(LR) `
`n_(N_(2)O_(5))/(1)xx0.4=n_(O_(3))=n_(O_(3))` formed `=n_(NO_(2))` formed/2
mole 16 22
`n_(NO_(2))+0.5=n_(O_(3))` formed `=16xx0.5=8` mole
`n_(O_(2))` used `=n_(NO_(2))` used `=16xx0.5=8` mole
`n_(O_(3))` total form =8+8=16
`n_(O_(2))+n_(O_(3))` after reaction =14+16=30