part 3: mol 02 remaining part4: Co mole fraction A mixture of 0.159 moles of C i
ID: 701004 • Letter: P
Question
part 3: mol 02 remaining
part4: Co mole fraction
A mixture of 0.159 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. The limiting reagent of the below reaction is carbon. 3C(s) + 202(g) CO2(g) +2CO(g) For 0.159 moles of carbon, determine the amounts of products (both the CO and CO2) formed in this reaction. Also determine the amount of O2 remaining and the mole fraction for CO. 1st attempt Part 1 (1 pt) td See Periodic Table See Hint mol CO2 formed Part 2 (1 pt) See Hint mol CO formedExplanation / Answer
From the stoichiometry of the reaction
3 moles of C produces = 1 mol of CO2 and 2 moles of CO
0.159 moles of C produces = 0.159/3 = 0.053 mol of CO2 and 2*0.159/3 = 0.106 moles of CO
And
3 moles of C reacts with = 2 mol of O2
0.159 moles of C reacts with = 2*0.159/3 = 0.106 mol of O2
Mol of O2 remaining = initial - reacted
= 0.117 - 0.106 = 0.011 mol
Total Moles at exit = mol O2 + mol CO + mol CO2
= 0.011 + 0.106 + 0.053
= 0.170
Mol fraction of CO = mol CO/total moles
= 0.106/0.170
= 0.6235
Mol CO2 formed = 0.053
Mol CO formed = 0.106
Mol O2 remains = 0.011
CO mol fraction = 0.6235
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