Question
A palladium or platinum catalyst was used in an automobile to convert carbon monoxide gas to carbon dioxide according to the following reaction:
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
A chemist researching the effectiveness of a new catalyst combines a 2.0 : 1.0 mole ratio mixture of carbon monoxide and oxygen gas (respectively) over the catalyst in a 2.45 L flask at a total pressure of 745 torr and a temperature of 552 K. When the reaction is complete, the pressure in the flask has dropped to 552 torr.
What percentage of the carbon monoxide was converted to carbon dioxide?
(a) 67.7 %
(b) 85.7 %
(c) 77.5 %
(d) 57.8 %
(e) 46.5 %
My attempt
I calculated the moles of gas in the original system
$$n = \frac{PV}{RT} = \frac{745 \times 2.45}{8.31 \times 552} = 0.398 \, \mathrm{mol}$$
Then I worked out the moles of $\ce{CO}$ and $\ce{O2}$ in the original system using the mole fractions.
$$\ce{2CO + O2 -> 2CO2}$$
moles of $\ce{CO}$ = $(2/3) \times 0.398 = 0.265\, \mathrm{mol}$
moles of $\ce{O2} = (1/3) \times 0.398 = 0.1326\,\mathrm{mol}$
moles of $\ce{CO2} = 0\, \mathrm{mol}$
Then I calculated the number of moles in the new system.
$$n-\frac{PV}{RT} = \frac{552 \times 2.45}{8.31 \times 552} = 0.295\, \mathrm{mol}$$
Then I worked out the moles of $\ce{CO, O2}$ and $\ce{CO2}$ in the new system using the mole fractions.
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
moles of $\ce{CO} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$
moles of $\ce{O2} = (1/5) \times 0.295 = 0.059\, \mathrm{mol}$
moles of $\ce{CO2} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$
Therefore percentage of CO reacted
$$\%_{\ce{CO}} = \frac{0.265 - 0.118}{0.265} \times 100 = 55.5\, \%$$
The correct answer according to the mark scheme is C) 77.5%. Is there any other method to get this answer?