Question

If 41.0 g of O2 and 5.6 g of CO2 are placed in a 6.8 L...

If 41.0 g of O2 and 5.6 g of CO2 are placed in a 6.8 L container at 25 oC , what is the pressure of the mixture of gases?


(-----------) atm

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Answer #1

Molar mass of O2 = 32 g/mol

mass(O2)= 41.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(41 g)/(32 g/mol)

= 1.281 mol

Molar mass of CO2,

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

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 5.6 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(5.6 g)/(44.01 g/mol)

= 0.1272 mol

We have:

total mol of gas = 1.281 mol + 0.1272 mol

= 1.408 mol

V = 6.8 L

T = 25.0 oC

= (25.0+273) K

= 298 K

use:

P * V = n*R*T

P * 6.8 L = 1.408 mol* 0.08206 atm.L/mol.K * 298 K

P = 5.0634 atm

Answer: 5.1 atm

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