Question

A mixture of .158 moles of C is reacted with .117 moles of O2 in a sealed, 10.0L vessel at 500K, producing a mixture of Co and CO2. The total pressure os .6961atm. What is the partial pressure of CO?

3 C(s) + 2 O2 (g) = 2 CO (g) + CO2 (g)

Answer 1

Since 3 moles of C reacts with 2 moles of O2,
0.158 moles of C will react with 2*0.158/3 = 0.105 moles of O2

So C is limiting reagent

Moles of CO formed = 2*0.158/3 = 0.105 moles
Moles of CO2 formed = 0.158/3 = 0.053 moles

Moles of O2 remaianing = 0.117 - 0.105 = 0.012 mol

3 C(s) + 2 O2 (g) ------> 2 CO (g) + CO2 (g)
0.158        0.117                       0                   0     (initial)
   0              0.012                    0.105            0.053   (final)

Total moles after reaction = 0.105 + 0.053 + 0.012 = 0.17 mol

Partial pressure of CO =moles of CO * total pressure / total moles
      = 0.105 * 0.6961 / 0.17
      = 0.4299 atm
Answer: 0.4299 atm