When 16 , g of oxygen and 3 , g of hydrogen a | vedclass

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

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$44.8 \, L$

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(C) $1$. Calculate the moles of each gas: Moles of $O_2 = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$. Moles of $H_2 = \frac{3 \, g}{2 \, g/mol} = 1.5 \, mol$. $2$. Total moles of the mixture: $n_{total} = 0.5 + 1.5 = 2.0 \, mol$. $3$. Use the Ideal Gas Law at $STP$ ($P = 1 \, atm = 760 \, mm \, Hg$,$T = 273 \, K$): $PV = nRT \implies V = \frac{nRT}{P}$. Since $1 \, mol$ of any ideal gas occupies $22.4 \, L$ at $STP$: $V = 2.0 \, mol 	imes 22.4 \, L/mol = 44.8 \, L$. $4$. Convert to $mL$: $44.8 \, L = 44800 \, mL$.

When $16 \, g$ of oxygen and $3 \, g$ of hydrogen are mixed at $760 \, mm$ pressure and $0 \, ^\circ C$,the total volume of the mixture will be approximately:

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