This set of Basic Chemical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Problem Solving Strategy”.
1. Solution-1 at the rate F combines with solution-2 at the rate D, their combination gives the product at the rate P, what is the equation of material balance?
a) F + D = P
b) F – P = D
c) D – F = P
d) None of the mentioned
View answer
Explanation: The general equation of material balance is F + D = P.
2. Solution-1 containing 30% sulfuric acid flowing at the rate 10 Kg/min combines with Solution-2 containing 20% sulfuric acid flowing at the rate 5 Kg/min, if their product is out at the rate 20 Kg/min what is the percentage of sulfuric acid in the product?
a) 10%
b) 20%
c) 30%
d) 40%
View answer
Explanation: Equation of material balance, 0.3(10) + 0.2(5) = x(20), => x = 0.2, => Percentage of sulfuric acid in the product = 20%.
3. Solution-1 containing 10% nitric acid flowing at the rate 10 Kg/min combines with Solution-2 containing 40% nitric acid flowing at the rate 5 Kg/min, if their product contains 30% nitric acid, what is the flow rate of product?
a) 10 Kg/min
b) 20 Kg/min
c) 30 Kg/min
d) 40 Kg/min
View answer
Explanation: Equation of material balance, 0.1(10) + 0.4(5) = 0.3(x), => x = 10 Kg/min.
4. Solution-1 containing 10% nitric acid and 20% acetic acid at the rate F combines with Solution-2 containing 20% nitric acid and 10% acetic acid at the rate D, the product formed contains 30% nitric acid and 20% acetic acid at the rate P, how many independent equations containing F, D, and P are possible?
a) 1
b) 2
c) 3
d) 4
View answer
Explanation: The equations of material balance are: 0.1F + 0.2D = 0.3P, 0.2F + 0.1D = 0.2P, and F + D = P.
5. Solution-1 containing 10% hydrochloric acid and 20% acetic acid at the rate F combines with Solution-2 containing 20% hydrochloric acid and 30% acetic acid at the rate D, the product formed contains 20% hydrochloric acid and 30% acetic acid at the rate P, how many independent equations containing F, D, and P are possible?
a) 1
b) 2
c) 3
d) 4
View answer
Explanation: The equations of material balance are: 0.1F + 0.2D = 0.2P, 0.2F + 0.3D = 0.3P and F + D = P, but if we subtract first two of them then we get the third, so number of independent equations are 2.
6. Solution-1 containing 40% nitric acid and 20% carbon dioxide at the rate F combines with Solution-2 containing 30% nitric acid and 40% carbon dioxide at the rate D, the product formed contains 30% nitric acid and 20% carbon dioxide at the rate P, If P = xF what is x?
a) 1
b) 2
c) 3
d) 4
View answer
Explanation: The equations of material balance are: 0.4F + 0.2D = 0.3P, 0.2F + 0.4D = 0.3P, and F + D = P, Solving these we get D = F = P/2, => x = 2.
7. Solution-1 containing 40% acetic acid and 20% carbon dioxide at the rate F combines with Solution-2 containing 20% acetic acid and 40% carbon dioxide at the rate D, the product formed contains 30% acetic acid at the rate P, what is the percentage of carbon dioxide in product?
a) 10%
b) 20%
c) 30%
d) 40%
View answer
Explanation: The equations for material balance are: 0.4F + 0.2D = 0.3P, F + D = P, and 0.2F + 0.4D = yP, solving these we get y = 0.3, => percentage of carbon dioxide = 30%.
8. Solution-1 containing 20% w1 and w2 at the rate 10 Kg/min combines with Solution-2 containing 10% w1, 20% w2 and w3 at the rate 20 Kg/min, the product formed contains and 10% w1, 30% w2 and 20% w3, what is the percentage of w2 in solution-1?
a) 5%
b) 10%
c) 25%
d) 50%
View answer
Explanation: Rate of product, P = 10 + 20 = 30 Kg/min, equation of material balance for w2: x(10) + 0.2(20) = 0.3(30), => w2 = 0.5, => percentage of w2 = 50%.
9. CH4 at the rate 10 mole/min is reacted with O2 at the rate 10 mole/min, what is the rate of formation of CO2 in the product?
a) 5 mole/min
b) 10 mole/min
c) 20 mole/min
d) 40 mole/min
View answer
Explanation: The balanced chemical reaction is: CH4 + 2O2 -> CO2 + 2H2O, this means O2 is the limiting reagent, => rate of formation of CO2 = 10/2 = 5 mole/min.
10. CH4 at the rate 10 mole/min is reacted with O2 at the rate 30 mole/min, what is the rate of formation of CO2 in the product?
a) 5 mole/min
b) 10 mole/min
c) 20 mole/min
d) 40 mole/min
View answer
Explanation: The balanced chemical reaction is: CH4 + 2O2 -> CO2 + 2H2O, this means CH4 is the limiting reagent, => rate of formation of CO2 = 10 mole/min.
11. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate F combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate D and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate P, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined
View answer
Explanation: The equations for material balance are: 0.3F + 0.2D + 0.1N = 0.2P, F + D + N = P, and 0.2F + 0.1D + 0.3N = yP, since there are 5 variables and 3 equation, the equations cannot be solved.
12. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate F combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate D and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate P, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined
View answer
Explanation: The equations for material balance are: 0.3F + 0.2D + 0.1N = 0.2P, F + D + N = P, and 0.2F + 0.1D + 0.3N = yP, since there are 5 variables and 3 equation, the equations cannot be solved.
13. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate F combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate D and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate 10 Kg/min, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined
View answer
Explanation: The equations for material balance are: 0.3F + 0.2D + 0.1N = 0.2(10), F + D + N = 10, and 0.2F + 0.1D + 0.3N = y(10), since there are 4 variables and 3 equation, the equations cannot be solved.
14. Solution-1 containing 30% acetic acid and 100% nitrogen at the rate 10 Kg/min combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate 20 Kg/min and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate 30 Kg/min the product formed contains 20% acetic acid, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined
View answer
Explanation: The equations for material balance are: P = 10 + 20 + 30 = 60 Kg/min, and 0.1(10) + 0.1(20) + 0.3(30) = y(60), => y = 0.1, => Percentage of nitrogen in the product = 10%.
15. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate 10 Kg/min combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate 20 Kg/min and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate P, what is the percentage of nitrogen in product?
a) 10%
b) 17.5%
c) 25.5%
d) 32.5%
View answer
Explanation: The equations for material balance, P = F + D + N = 30 + N, and 0.3(10) + 0.2(20) + 0.1N = 0.2P = 0.2(30 + N), => N = 10 Kg/min, => P = 40 Kg/min, equation of material balance for nitrogen: 0.2(10) + 0.1(20) + (0.3)10 = y(40), => y = 7/40 = 0.175, => Percentage of nitrogen in the product is 17.5%.
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