Question

A tank contains a mixture of 48.2 g of oxygen gas and 70.0 g of carbon dioxide gas at 23 °C. The total pressure in the tank is 8.40 atm. Calculate the partial pressure (in atm) of each gas in the mixture. Partial pressure = atm O2 Partial pressure = atm CO2

Answer 1

1)

Molar mass of O2 = 32 g/mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

n(O2) = mass of O2/molar mass of O2

= 48.2/32

= 1.5063

n(CO2) = mass of CO2/molar mass of CO2

= 70.0/44.01

= 1.5905

n(O2),n1 = 1.5063 mol

n(CO2),n2 = 1.5905 mol

Total number of mol = n1+n2

= 1.5063 + 1.5905

= 3.0968 mol

Partial pressure of each components are

p(O2),p1 = (n1*Ptotal)/total mol

= (1.5063 * 8.4)/3.0968

= 4.0857 atm

p(CO2),p2 = (n2*Ptotal)/total mol

= (1.5905 * 8.4)/3.0968

= 4.3143 atm

Answer:

p(O2) = 4.09 atm

p(CO2) = 4.31 atm

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