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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.
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Solution
`"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
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