$(a)\;76\%\qquad(b)\;75\%\qquad(c)\;74\%\qquad(d)\;78\%$

$2H_2 + O_2\rightarrow 2H_2O(l)$

Volume before reaction of $H_2O$ = a

Volume before reaction of $O_2$ = b

Volume before reaction of $2H_2O$ = O

Volume after reaction of $H_2O$ = (a-2b)

Volume after reaction of $O_2$ = O

Since at constnt P and T gases react in their volume ratio

a + b = 40

a - 2b = 10

$\therefore$ a = 30mL

b = 10 mL

Mole $\% of H_2 = Volume \% of H_2 = \large\frac{30}{40}\times100$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= 75$

Hence answer is (b)

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