16 , g of oxygen and 3 , g of hydrogen are mi | vedclass

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(C) Step $1$: Calculate the number of moles of each gas.$n(O_2) = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$$n(H_2) = \frac{3 \, g}{2 \, g/mol} = 1.5 \

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$44.8 \, L$

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(C) Step $1$: Calculate the number of moles of each gas.$n(O_2) = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$$n(H_2) = \frac{3 \, g}{2 \, g/mol} = 1.5 \, mol$Step $2$: Calculate the total number of moles.$n_{total} = 0.5 \, mol + 1.5 \, mol = 2.0 \, mol$Step $3$: Use the ideal gas equation $PV = nRT$.Given $P = 760 \, mm \, Hg = 1 \, atm$,$T = 0 \, ^oC = 273 \, K$,and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.$V = \frac{nRT}{P} = \frac{2.0 \, mol 	imes 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 273 \, K}{1 \, atm} \approx 44.8 \, L$.Since $1 \, L = 1000 \, mL$,$44.8 \, L = 44800 \, mL$.

$16 \, g$ of oxygen and $3 \, g$ of hydrogen are mixed and kept at $760 \, mm$ pressure and $0 \, ^oC$. The total volume occupied by the mixture will be nearly

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