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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. 16B. 21C. 27D. 30 |
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Answer» `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 |
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