At T(K),a gaseous mixture contains H_2 and O_ | vedclass

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(B) Given: Total pressure $P_{total} = 2 \ bar$,Partial pressure of $H_2$ $(P_{H_2})$ = $1.778 \ bar$. Partial pressure of $O_2$ $(P_{O_2})$ = $P_{tot

Vedclass · Accepted Answer

$33.33$

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(B) Given: Total pressure $P_{total} = 2 \ bar$,Partial pressure of $H_2$ $(P_{H_2})$ = $1.778 \ bar$. Partial pressure of $O_2$ $(P_{O_2})$ = $P_{total} - P_{H_2} = 2 - 1.778 = 0.222 \ bar$. According to Dalton's Law,the mole fraction $(x)$ is proportional to the partial pressure: $x_{H_2} = \frac{P_{H_2}}{P_{total}} = \frac{1.778}{2} = 0.889$ and $x_{O_2} = \frac{P_{O_2}}{P_{total}} = \frac{0.222}{2} = 0.111$. Assuming $1 \ mole$ of the mixture,moles of $H_2$ $(n_{H_2})$ = $0.889 \ mol$ and moles of $O_2$ $(n_{O_2})$ = $0.111 \ mol$. Mass of $H_2$ = $n_{H_2} 	imes M_{H_2} = 0.889 	imes 2 = 1.778 \ g$. Mass of $O_2$ = $n_{O_2} 	imes M_{O_2} = 0.111 	imes 32 = 3.552 \ g$. Total mass = $1.778 + 3.552 = 5.33 \ g$. Weight percentage of $H_2$ = $(\frac{1.778}{5.33}) 	imes 100 \approx 33.33 \%$.

At $T(K)$,a gaseous mixture contains $H_2$ and $O_2$. The total pressure of the mixture is $2 \ bar$. The partial pressure of $H_2$ is $1.778 \ bar$. What is the weight $(w/w)$ percentage of $H_2$ in the mixture?

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