Question

A mixture of 0.154 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L vessel at 500.0 K, producing a mixture of CO and CO2.

3C(s)+2O2​(g) ---> 2CO(g)+CO2​(g)

The total pressure is 0.691 atm. What is the partial pressure of CO?

Answer 1

Balanced chemical equation is:
3 C + 2 O2 ---> 2 CO + CO2


3 mol of C reacts with 2 mol of O2
for 0.154 mol of C, 0.1027 mol of O2 is required
But we have 0.117 mol of O2

so, C is limiting reagent
we will use C in further calculation\

According to balanced equation
mol of CO formed = (2/3)* moles of C
= (2/3)*0.154
= 0.1027 mol

According to balanced equation
mol of CO2 formed = (1/3)* moles of C
= (1/3)*0.154
= 5.133*10^-2 mol

According to balanced equation
mol of O2 reacted = (2/3)* moles of C
= (2/3)*0.154
= 0.1027 mol
mol of O2 remaining = mol initially present - mol reacted
mol of O2 remaining = 0.117 - 0.1027
mol of O2 remaining = 0.0143 mol

partial pressure of CO = moles of CO * total pressure / total moles
= 0.1027*0.691/(0.1027 + 5.133*10^-2 + 0.0143)
= 0.1027*0.691/0.16833
= 0.422 atm

Answer: 0.422 atm