Question

Question #4a0c4

Answer 1

`7.2"g of "H_2O`

Explanation:

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`