15 Question (4 points) e See page 459 Amixture of 0.145 moles of C is reacted wi

Question

15 Question (4 points) e See page 459 Amixture of 0.145 moles of C is reacted with 0.117 moles of O2 inasealed, 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 0.145 moles of carbon, determine the amounts of products (both the CO and CO2) formed in this reaction. Also determine the amount of Oz remaining and the mole fraction for CO 3rd attempt Part 1 (1 point) See Periodic Table Q See Hint mol CO2 formed Part 2 (1 point) See Hint 096 mol CO formed Part 3 (1 point) See Hint 0585 mol O2 remaining Part 4 (1 point) See Hint CO mole fraction: 15,16 > 13 OF16QUESTIONS COMPLETED + VIEw SOLUTIO SUBMIT ANSWER

Explanation / Answer

3C + 2O2 ------------> CO2 + 2CO

1) according to balanced reaction

3 moles C gives 1 mole CO2

0.145 moles C gives = 0.145 x 1 / 3 = 0.0483 moles CO2

moles of CO2 formed = 0.0483

2) 3 moles C forms 2 moles CO

0.145 moles C forms 0.145 x 2 / 3 = 0.097 moles

moles of CO formed = 0.097

3) 3 moles C reacts with 2 moles O2

0.145 moles C reacts with 0.145 x 2 / 3 = 0.097

moles of O2 left = 0.097

4) mole fraction of CO = moles of CO / total moles

moles of CO = 0.097

total moles = 0.0483 + 0.097 + 0.097 = 0.2423

mole fraction of CO = 0.097 / 0.2423

mole fraction of CO = 0.40

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