Q2-Q6. Two moles ofpropane (C-H8) gas is burned with 20% excess air Molecular we

Question

Q2-Q6. Two moles ofpropane (C-H8) gas is burned with 20% excess air Molecular weights: C: 12 0:16 N: 14 2. The % of CO2 in the flue gas on wet basis is most nearly (v/v): a. 5% b. 10% c. 15% d. 2090 e. 25% f. other, write your answer 3. The % of CO2 in the flue gas on dry basis (v/v) 690 a. b. 11% c. 16% d. 2190 e. 26% f other, write your answer 4. The % of CO2 in the flue gas on wet basis is most nearly (w/w): a. b. c. 5% 10% 15% d. e. f. 2090 21% other, write your answer 5. Volume of propane burned at STP is most nearly: 10.4 L 22.4 L a. b. c. 44.8 L d. 50.2 L e. 67.2 L . other, write your answer

Explanation / Answer

Combustion of Two moles of C3H8

2C3H8 + 10 O2 = 6 CO2 + 8 H2O

Theoretical O2 required = 10 mol O2 / 2 mol C3H8

Theoretical air required = 10/0.21

= 47.62 mol air/2 mol C3H8

Actual air supplied = 1 .20 x 47.62

= 57.144 mol air/2 mol C3H8

Moles of O2 supplied = 57.144*0.21

= 12 mol of O2/2 mol C3H8

Moles of N2 supplied = 45.144 mol/2 mol C3H8

In the outlet

Moles of CO2 = 6

Moles of H2O = 8

Moles of C3H8 = 0

Moles of O2 = 12 - 10 = 2

Moles of N2 = 45.144

Ans 2

Total Moles wet basis = 61.144

Mol% CO2 wet basis = 6*100/61.144 = 9.812% = 10%

Option B is the correct answer

Ans 3

Total Moles dry basis = 53.144

Mol% CO2 wet basis = 6*100/53.144 = 11.29% = 11%

Option B is the correct answer

Ans 4

Mass of CO2 = 6*44 = 264 g

Mass of H2O = 8*18 = 144 g

Mass of C3H8 = 0

Mass of O2 = 12 - 10 = 2 *32 = 64 g

Mass of N2 = 45.144*28 = 1264.032 g

Total mass wet basis = 1736.032 g

% CO2 wet basis = 264*100/1736.032 = 15.2% = 15%

Option C is the correct answer

Ans 5

At STP

Pressure P = 1 atm

Temperature T = 273 K

Volume = nRT/P

= 2*0.0821*273/1

= 44.8266 L

Option C is the correct answer

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