This set of Basic Chemical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Material Balances Involving Combustion”.
1. What is the percentage of excess air, if 10 moles of air entered the process and only 5 moles of that are required?
a) 10%
b) 50%
c) 75%
d) 100%
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Explanation: Percentage of excess air = 100*(10 – 5)/5 = 100%.
2. What is the percentage of excess air if 50 grams of air is in excess and 150 grams of air enters the process?
a) 10%
b) 25%
c) 50%
d) 100%
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Explanation: Percentage of excess air = 100*50/(150 – 50) = 50%.
3. 10 moles of ethane is supplied with 49 moles of oxygen, what is the percentage of excess oxygen?
a) 20%
b) 40%
c) 50%
d) 70%
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Explanation: The balanced reaction is C2H6 + 3.5O2 -> 2CO2 + 3H2O, => O2 required = 3.5*10 = 35 moles, => percentage of excess oxygen = 100*(49 – 35)/35 = 40%.
4. The extent of a reaction is 10, and the stoichiometric coefficient of O2 is 2.5, if the moles of O2 entering the process is 45, what is the percentage of excess air?
a) 20%
b) 40%
c) 60%
d) 80%
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Explanation: Moles of O2 required = 10*2.5 = 25, => Percentage of excess air = 100*(45 – 25)/25 = 80%.
5. 144 grams of C5H12 is burnt with 64 grams O2, and 44 grams of CO2 is formed, what is the percentage of excess O2?
a) 25%
b) 50%
c) 75%
d) 100%
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Explanation: The balanced chemical reaction is C5H12 + 8O2 -> 5CO2 + 6H2O, => Extent of reaction = (44/44 – 0)/5 = 0.2. Moles of O2 entered the process = 64/32 = 2, Moles of O2 required = 0.2*8 = 1.6, => Percentage of excess oxygen = 100*(2 – 1.6)/1.6 = 25%.
6. Ethene is burnt with 50% of excess air, what is the percentage of CO2 in the products?
a) 11.11%
b) 36.36%
c) 66.66%
d) 72.72%
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Explanation: The reaction is C2H4 + 3O2 -> 2CO2 + 2H2O. Basis: 10 moles of C2H4, => moles of O2 reacted = 30, => moles of O2 entered the process = 45, => moles of CO2, H2O and O2 in products are 20, 20 and 15 respectively, => percentage of CO2 = 20/55*100 = 36.36%.
7. Ethene is burnt with 150% of excess air, what is the percentage of O2 in the products?
a) 12.5%
b) 33.3%
c) 45.4%
d) 52.9%
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Explanation: The reaction is C2H4 + 3O2 -> 2CO2 + 2H2O. Basis: 10 moles of C2H4, => moles of O2 reacted = 30, => moles of O2 entered the process = 75, => moles of CO2, H2O and O2 in products are 20, 20 and 45 respectively, => percentage of O2 = 45/85*100 = 52.9%.
8. Pentane is burnt with 100% of excess air, what is the percentage of H2O in the products?
a) 12.4%
b) 34.6%
c) 42.1%
d) 56.9%
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Explanation: The reaction is C5H12 + 8O2 -> 5CO2 + 6H2O. Basis: 10 moles of C5H12, => moles of O2 reacted = 80, => moles of O2 entered the process = 160, => moles of CO2, H2O and O2 in products are 50, 60 and 80, => percentage of H2O = 80/190*100 = 42.1%.
9. Propane is burnt with 20% excess O2, what is the percentage of CO2 in products?
a) 12.5%
b) 25%
c) 45%
d) 75.5%
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Explanation: The reaction is C3H8 + 5O2 -> 3CO2 + 4H2O. Basis: 10 moles of C3H8, => moles of O2 reacted = 50, => moles of O2 entered the process = 60, => moles of CO2, H2O and O2 in products are 30, 40 and 10, => percentage of CO2 = 10/80*100 = 12.5%.
10. Octane is burnt with 40% excess O2, what is the percentage of CO2 in products?
a) 18.18%
b) 36.36%
c) 54.54%
d) 72.72%
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Explanation: The reaction is C8H18 + 12.5O2 -> 8CO2 + 9H2O. Basis: 10 moles of C8H18, => moles of O2 reacted = 125, => moles of O2 entered the process = 175, => moles of CO2, H2O and O2 in products are 80, 90 and 50, => percentage of CO2 = 80/220*100 = 36.36%.
11. Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the rate of products?
a) 900 moles
b) 950 moles
c) 1000 moles
d) 1050 moles
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Explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Rate of products = 790 + 110 + 50 + 100 = 1050 moles.
12. Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of O2 in products?
a) 10.4%
b) 25.8%
c) 45.6%
d) 78.2%
View answer
Explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Percentage of O2 = 110/1050*100 = 10.4%.
13. Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of CO2 in products?
a) 4.7%
b) 10.4%
c) 16.5%
d) 24.3%
View answer
Explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Percentage of CO2 = 50/1050*100 = 4.7%.
14. Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of N2 in products?
a) 25.4%
b) 45.2%
c) 75.2%
d) 95.6%
View answer
Explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => O2 unreacted = 210 – 100 = 110 moles, CO2 formed = 50 moles, H2O formed = 100 moles. => Percentage of N2 = 790/1050*100 = 75.2%.
15. Methane is at the rate 50 moles is supplied to a reactor with air at the rate 1000 moles, the reactor leaves O2, N2, CO2 and H2O, what is the percentage of H2O in products?
a) 5.8%
b) 9.5%
c) 15.4%
d) 20.7%
View answer
Explanation: The reaction is CH4 + 2O2 -> CO2 + 2H2O. O2 supplied = 1000*21/100 = 210 moles, N2 supplied = 1000*79/100 = 790 moles, => Osub>2 unreacted = 210 – 100 = 110 moles, COsub>2 formed = 50 moles, Hsub>2O formed = 100 moles. => Percentage of H2O = 100/1050*100 = 9.5%.
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