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.

For the following reaction : N_(2)O_(5)(g)+O_(2)(g)overset(40%)rarr2N

1.	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.
Answer» 16 21 27 30 Solution :`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