Question

A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO₂ at 300 K the total pressure in the tanks is 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.

Answer 1

`"m"_("O"_2)` = 52.5 g

`"P"_("O"_2)` = ?

`"m"_("CO"_2)` = 65.1 g

`"P"_("CO"_2)` = ?

T = 300 K

P = 9.21 atm

`"P"_("O"_2) = "X"_("O"_2) xx "Total Pressure"`

`"X"_("O"_2) = "n"_("O"_2)/("n"_("O"_2) + "n"_("CO"_2))`

`"n"_("O"_2) = "Mass of O"_2/("Molar mass of O"_2)`

= `(52.5  "g")/(32  "g mol"^-1)`

= 1.64 mol

`"n"_("CO"_2) = "Mass of CO"_2/"Molar mass of CO"_2`

= `(65.1  "g")/(44  "g mol"^-1)`

= 1.48 mol

`"X"_("O"_2) = "n"_("O"_2)/("n"_("O"_2) + "n"_("CO"_2))`

= `1.64/3.12`

= 0.53

`"X"_("CO"_2) = "n"_("CO"_2)/("n"_("O"_2) + "n"_("CO"_2))`

= `1.48/3.12`

= 0.47

`"P"_("O"_2) = "X"_("O"_2) xx "Total Pressure"`

= 0.53 × 9.21 atm

= 4.88 atm

`"P"_("CO"_2) = "X"_("CO"_2) xx "Total Pressure"`

= 0.47 × 9.21 atm

= 4.33 atm