Question #4a0c4

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Question #4a0c4

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Answer 1 · 2017-05-15T17:11:38

$7.2 g of H_{2} O$ $7.2"g of "H_2O$

Explanation:

Use the Universal Gas Law with $R = 62.364 \frac{L∙torr}{mol∙K}$ $R = 62.364" L•torr"/"mol•K"$

$P V = n R T$ $PV=nRT$

$n = \frac{P V}{R T}$ $n = (PV)/(RT)$

Given $P = 512 torr, V = 8.22 L, and T = 322^{\circ} K$ $P = 512"torr", V = 8.22"L, and "T= 322^@"K"$

$n = \frac{(512 torr) (8.22 L)}{(62.364 \frac{L∙torr}{mol∙K}) (322^{\circ} K)}$ $n = ((512"torr")(8.22"L"))/((62.364" L•torr"/"mol•K")(322^@"K"))$

$n = 0.20 mol of O_{2}$ $n = 0.20"mol of "O_2$

Use Dimensional Analysis (Factor-Label Method):

$\frac{0.20 mol of O_{2}}{1} \frac{2 mol of H_{2} O}{1 mol of O_{2}} \frac{18 g of H_{2} O}{1 mol of H_{2} O} = 7.2 g of H_{2} O$ $(0.20"mol of "O_2)/1(2"mol of "H_2O)/(1"mol of "O_2)(18"g of "H_2O)/(1"mol of "H_2O) = 7.2"g of "H_2O$

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Question #4a0c4

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