Question

A mixture of CO(g) and O2(g) in a 1.1 −L container at 1.0×103 K has a total pressure of 2.3 atm . After some time the total pressure falls to 1.8 atm as the result of the formation of CO2.

Find the mass (in grams) of CO2 that forms.

Answer 1

The reaction taking place is:

2CO + O2 —> 2CO2

We can see from reaction that when 2 mol of CO2 is formed, reduction is total mol is by 1.

That is mol (or pressure) of CO2 formed = 2*reduction in mol (or pressure)

Here,

reduction in pressure = 2.3 - 1.8 = 0.5 atm

So,

pressure of CO2 formed = 2*0.5 atm

= 1.0 atm

Given:

P = 1.0 atm

V = 1.1 L

T = 1000.0 K

find number of moles using:

P * V = n*R*T

1 atm * 1.1 L = n * 0.08206 atm.L/mol.K * 1000 K

n = 1.34*10^-2 mol

Molar mass of CO2,

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

= 1*12.01 + 2*16.0

= 44.01 g/mol

use:

mass of CO2,

m = number of mol * molar mass

= 1.34*10^-2 mol * 44.01 g/mol

= 0.5899 g

Answer: 0.590 g