16.8 L of a gas mixture containing H_2 and O | vedclass

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(B) The electrolysis of water is represented by the equation: $2H_2O(l) 	o 2H_2(g) + O_2(g)$.The total moles of gas produced is $n_{total} = \frac{16

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The electrolysis of water is represented by the equation: $2H_2O(l) 	o 2H_2(g) + O_2(g)$.The total moles of gas produced is $n_{total} = \frac{16.8 \ L}{22.4 \ L/mol} = 0.75 \ mol$.According to the stoichiometry,$2 \ mol$ of $H_2$ and $1 \ mol$ of $O_2$ are produced from $2 \ mol$ of $H_2O$. Thus,the ratio of $H_2:O_2$ is $2:1$.Let the moles of $O_2$ be $x$. Then the moles of $H_2$ are $2x$.$x + 2x = 0.75 \implies 3x = 0.75 \implies x = 0.25 \ mol$ of $O_2$.Therefore,moles of $H_2 = 0.5 \ mol$.From the balanced equation,$2 \ mol$ of $H_2$ are produced from $2 \ mol$ of $H_2O$,so $0.5 \ mol$ of $H_2$ are produced from $0.5 \ mol$ of $H_2O$.Mass of $H_2O = 0.5 \ mol 	imes 18 \ g/mol = 9 \ g$.

$16.8 \ L$ of a gas mixture containing $H_2$ and $O_2$ is formed at $NTP$ by the electrolysis of water. What is the mass of the electrolysed water in $g$?

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