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% w_{1} and w_{2} at the rate 10 Kg/min combines with Solution-2 containing 10% w_{1}, 20% w_{2} and w_{3} at the rate 20 Kg/min, the product formed contains and 10% w_{1}, 30% w_{2} and 20% w_{3}, what is the percentage of w_{2} 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 w

_{2}: x(10) + 0.2(20) = 0.3(30), => w

_{2}= 0.5, => percentage of w

_{2}= 50%.

9. CH_{4} at the rate 10 mole/min is reacted with O_{2} at the rate 10 mole/min, what is the rate of formation of CO_{2} 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: CH

_{4}+ 2O

_{2}-> CO

_{2}+ 2H

_{2}O, this means O

_{2}is the limiting reagent, => rate of formation of CO

_{2}= 10/2 = 5 mole/min.

10. CH_{4} at the rate 10 mole/min is reacted with O_{2} at the rate 30 mole/min, what is the rate of formation of CO_{2} 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: CH

_{4}+ 2O

_{2}-> CO

_{2}+ 2H

_{2}O, this means CH

_{4}is the limiting reagent, => rate of formation of CO

_{2}= 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%.

**Sanfoundry Global Education & Learning Series – Basic Chemical Engineering.**

To practice all areas of Basic Chemical Engineering, __here is complete set of 1000+ Multiple Choice Questions and Answers__.