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Answer 1 · 2017-05-15T17:11:38

$7.2"g of "H_2O$

Use the Universal Gas Law with $R = 62.364" L•torr"/"mol•K"$

$PV=nRT$

$n = (PV)/(RT)$

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

$n = ((512"torr")(8.22"L"))/((62.364" L•torr"/"mol•K")(322^@"K"))$

$n = 0.20"mol of "O_2$

Use Dimensional Analysis (Factor-Label Method):

$(0.20"mol of "O_2)/1(2"mol of "H_2O)/(1"mol of "O_2)(18"g of "H_2O)/(1"mol of "H_2O) = 7.2"g of "H_2O$