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If 41.7 g of O2 and 4.5 g of CO2 are placed in a 8.1 L container at 29 °C, what is the pressure of the mixture of gases? AC *

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Answer #1

Molar mass of O2 = 32 g/mol

mass(O2)= 41.7 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(41.7 g)/(32 g/mol)

= 1.303 mol

Molar mass of CO2,

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

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 4.5 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(4.5 g)/(44.01 g/mol)

= 0.1022 mol

Total mol gas = 1.303 mol + 0.1022 mol

= 1.405 mol

Given:

V = 8.1 L

n = 1.405 mol

T = 29.0 oC

= (29.0+273) K

= 302 K

use:

P * V = n*R*T

P * 8.1 L = 1.405 mol* 0.08206 atm.L/mol.K * 302 K

P = 4.30 atm

Answer: 4.30 atm

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