Chemical Engineering

Element Material Balances

Question 1
Marks : +2 | -2
Pass Ratio : 50%
A reactor is supplied with 10 moles glucose and excess oxygen to produce CO2 and H2O, if the rate of products is 25 moles, what is the percentage of CO2 in the products?
0.1
0.2
0.3
0.4
Explanation:
The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O. Let the fraction of CO2 be x, Element balances, C: 10(6) = 6*x*5(1), => x = 0.4.
Question 2
Marks : +2 | -2
Pass Ratio : 50%
A reactor is supplied with 10 moles glucose and excess oxygen to produce 40% CO2, 60% H2O, what is the rate of products?
25 mole
50 mole
75 mole
100 mole
Explanation:
The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O, Let the rate of products be P. Element balances, C: 10(6) = 6*0.4P(1), => P = 25 mole.
Question 3
Marks : +2 | -2
Pass Ratio : 50%
A reactor supplied with CO2 and H2O, the product contains H2O and H2CO3, if the rate of H2O is 10 mole, what is the rate of products?
5 mole
10 mole
15 mole
20 mole
Explanation:
The reaction is CO2 + H2O -> H2CO3. Let the rate of products be P, and the fraction of H2CO3 in products be x, material balances, H: 10(2) = x*P(2) + (1 – x)*P(2), => P = 10 mole.
Question 4
Marks : +2 | -2
Pass Ratio : 50%
A reactor supplied with CO2 and H2O, the product contains 10% H2CO3 and 90% H2O, what is the feed rate of CO2, if the rate of product is 50 mole?
5 mole
10 mole
25 mole
50 mole
Explanation:
The reaction is CO2 + H2O -> H2CO3. Let the rates of CO2 and H2O be F and W respectively. Element balances, C: F(1) = 0.1P(1), => 10F = P, => F = 5 mole.
Question 5
Marks : +2 | -2
Pass Ratio : 50%
A reactor is supplied with 10 mole of Hexane and some hydrogen to produce 20% methane, 30% propane and 50% butane, what is the rate of products?
10 mole
16.6 mole
26.6 mole
40 mole
Explanation:
The reaction is 2C6H14 + 2H2 -> CH4 + 2C3H8 + C5H12, Let the rate of products be P. Element balance, C: 2*10(6) = 0.2P(1) + 2*0.3P(3) + 0.5P(5), => 120 = 4.5P, => P = 26.6 mole.
Question 6
Marks : +2 | -2
Pass Ratio : 50%
A reactor supplied with CO2 and H2O, the product contains H2O and H2CO3, if the rate of CO2 is 10 mole, what is the percentage of H2CO3 in products?
10
50
75
100
Explanation:
The reaction is CO2 + H2O -> H2CO3. Let the rate of CO2 and H2O and product be F and W respectively, and the fraction of H2CO3 in products be x. Element balances, C: F(1) = x*10(1), => F = 10x, => x = 1, => percentage of H2CO3 = 100%.
Question 7
Marks : +2 | -2
Pass Ratio : 50%
A reactor supplied with CO2 and H2O, the product contains CO2, H2O and H2CO3, if the rate of products is 10 mole and percentage of CO2 in products is 20%, what is the feed rate of H2O?
2 mole
4 mole
8 mole
10 mole
Explanation:
The reaction is CO2 + H2O -> H2CO3. Let the rate of H2O be W, and percentage of H2O in products be x. Material balances, H: W(2) = x*10(2) + (1 – x – 0.2)*10(2), => W = 8 mole.
Question 8
Marks : +2 | -2
Pass Ratio : 50%
A reactor is supplied with Hexane and hydrogen to produce 20% methane, 30% propane and 50% butane, what is the ratio of hydrogen consumed and the hexane reacted?
0.21
0.45
0.59
0.73
Explanation:
The reaction is 2C6H14 + 2H2 -> CH4 + 2C3H8 + C5H12, Let the rate of Hexane, hydrogen and products be F, W and P respectively. Element balances, C: 2F(6) = 0.2P(1) + 2*0.3P(3) + 0.5P(5), => 8F = 3P, H: 2F(14) + 2W(2) = 0.2P(4) + 2*0.3P(8) + 0.5P(12), => 7F + W = 2.9P, => W/F= 2.2/3 = 0.73.
Question 9
Marks : +2 | -2
Pass Ratio : 50%
A reactor supplied with CO2 and H2O, the product contains 40% CO2, 30% H2O and 30% H2CO3, what is the ratio of feed rate of CO2 and rate of products?
0.3
0.4
0.7
0.9
Explanation:
The reaction is CO2 + H2O -> H2CO3. Let the rate of CO2, H2O and H2CO3 be F, W and P respectively. Element balances, C: F(1) = 0.4P(1) + 0.3P(1), => F/P = 0.7.
Question 10
Marks : +2 | -2
Pass Ratio : 50%
A reactor is supplied with glucose and oxygen to produce 40% CO2, 60% H2O, what is the ratio of glucose reacted and products?
0.2
0.4
0.6
0.8
Explanation:
The reaction is C6H12O6 + 6O2 -> 6CO2 + 6H2O, Let the rate of glucose and products be F and P, Element balances, C: F(6) = 6*0.4P(1), => F/P = 0.4.