At 100,^oC,a 1.5,L vessel contains 0.4,g of O | vedclass

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(A) Step $1$: Calculate moles of gases.$n(O_2) = \frac{0.4\,g}{32\,g/mol} = 0.0125\,mol$.$n(H_2) = \frac{0.06\,g}{2\,g/mol} = 0.03\,mol$.Step $2$: Tot

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$0.867$

Vedclass · Answer

(A) Step $1$: Calculate moles of gases.$n(O_2) = \frac{0.4\,g}{32\,g/mol} = 0.0125\,mol$.$n(H_2) = \frac{0.06\,g}{2\,g/mol} = 0.03\,mol$.Step $2$: Total pressure using $PV = nRT$.$n_{total} = 0.0125 + 0.03 = 0.0425\,mol$.$P = \frac{nRT}{V} = \frac{0.0425 	imes 0.0821 	imes 373}{1.5} = 0.867\,atm$.Step $3$: Reaction $2H_2 + O_2 ightarrow 2H_2O$.$0.0125\,mol$ of $O_2$ reacts with $0.025\,mol$ of $H_2$ to form $0.025\,mol$ of $H_2O$.Remaining $H_2 = 0.03 - 0.025 = 0.005\,mol$.Partial pressure of $H_2 = \frac{0.005 	imes 0.0821 	imes 373}{1.5} = 0.102\,atm$.Partial pressure of $H_2O = \frac{0.025 	imes 0.0821 	imes 373}{1.5} = 0.51\,atm$.

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