In: Chemistry
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)
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