10 mol of a mixture of hydrogen and oxygen g | vedclass

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(B) $2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)}$$18 \ g$ of $H_2O = 1 \ mol$,so $3.6 \ g$ of $H_2O = \frac{3.6}{18} = 0.2 \ mol$.According to th

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$0.97$

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(B) $2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)}$$18 \ g$ of $H_2O = 1 \ mol$,so $3.6 \ g$ of $H_2O = \frac{3.6}{18} = 0.2 \ mol$.According to the stoichiometry,$0.2 \ mol$ of $H_2O$ is produced from $0.2 \ mol$ of $H_2$ and $0.1 \ mol$ of $O_2$.Total moles of gases consumed $= 0.2 + 0.1 = 0.3 \ mol$.Initial moles of gas mixture $= 10 \ mol$.Remaining moles of gas mixture $= 10 - 0.3 = 9.7 \ mol$.Since $V$ and $T$ are constant,$P \propto n$.$\frac{P_1}{n_1} = \frac{P_2}{n_2} \Rightarrow \frac{1 \ atm}{10 \ mol} = \frac{P_2}{9.7 \ mol}$.$P_2 = \frac{9.7}{10} = 0.97 \ atm$.

$10 \ mol$ of a mixture of hydrogen and oxygen gases at a pressure of $1 \ atm$ at constant volume and temperature,react to form $3.6 \ g$ of liquid water. The pressure of the resulting mixture will be closest to $..... \ atm$

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