A vessel contains 14 , g (7 moles) of hydroge | vedclass

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(A) Initial moles of $H_2$ $(n_{H_2})$ $= 14 \, g / 2 \, g/mol = 7 \, mol$.Initial moles of $O_2$ $(n_{O_2})$ $= 96 \, g / 32 \, g/mol = 3 \, mol$.The

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(A) Initial moles of $H_2$ $(n_{H_2})$ $= 14 \, g / 2 \, g/mol = 7 \, mol$.Initial moles of $O_2$ $(n_{O_2})$ $= 96 \, g / 32 \, g/mol = 3 \, mol$.The chemical reaction is: $2H_2 + O_2 ightarrow 2H_2O$.According to the stoichiometry,$2 \, mol$ of $H_2$ reacts with $1 \, mol$ of $O_2$.Therefore,$3 \, mol$ of $O_2$ will react with $6 \, mol$ of $H_2$.Since we have $7 \, mol$ of $H_2$,$O_2$ is the limiting reagent and will be completely consumed.Remaining moles of $H_2 = 7 - 6 = 1 \, mol$.The product $H_2O$ formed is in liquid state at $273 \, K$,so its contribution to the gas pressure is negligible.Using the ideal gas law $PV = nRT$,since $V$ and $T$ are constant,$P \propto n$.Initial pressure $P_i = 1 \, atm$ (at $STP$ with $n_i = 7 + 3 = 10 \, mol$).Final pressure $P_f = (n_f / n_i) 	imes P_i = (1 / 10) 	imes 1 \, atm = 0.1 \, atm$.

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