This set of Engineering Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Combustion Calculation”.
1. Which of the following law is not used in combustion calculation?
a) The law of conservation of mass
b) The law of definite proportion
c) The law of multiple proportion
d) The law of conservation of energy
Explanation: According to the law of multiple proportion, “If two elements combine together to form more than one compound, than the ratios of masses of the second element which combines with a fixed mass of first element will be ratios of small whole number”. Since combustion does not involve multiple elements, this law is not required.
2. Which of the following cannot be calculated in combustion calculation?
a) Calorific value of fuel
b) Composition of fuel
c) Composition of the flue gas
d) Nature of combustion
Explanation: The calorific value of coal can be calculated by using bomb calorimeter. If two of the following are given from composition of fuel, composition of the flue gas and nature of combustion than the third one can be calculated in the combustion.
3. Which of the following is the ideal gas equation for 1 mol of gas?
a) PV = nRT
b) PV = 6.022nRT
c) PR = nVT
d) PV = RT
Explanation: The ideal gas equation is as follows-
PV = nRT, since n=1
where P is pressure, V is volume, R is gas constant, T is temperature and n is the number of moles.
The value of R is 8.314 J/mol K.
4. What is the gram molecular volume of all gases at normal temperature pressure?
a) 2.24 L
b) 22.4 L
c) 224 L
d) 2240 L
Explanation: According to the Avogadro law, The gram molecular volume of all gases at N.T.P. (i.e., 0 0C and 760 mm abs) is 22.4 litres. The corresponding volume in the English system is 359 cu. Ft.
5. Which of the following laws can be used for calculating the volume of gas at a given temperature and pressure?
a) Boyle’s and Avogadro law
b) Charle’s and Avogadro law
c) Charle’s and Dalton’s law
d) Boyle’s and Charle’s law
Explanation: Boyle’s and Charle’s law can be used for reducing the volume of a gas at a given temperature and pressure to the corresponding volume at any other specific conditions of temperature and pressure with the help of the equation-
(PV)/T = (P1V1)/T1.
6. What is the absolute temperature of gases at N.T.P.?
c) -273 K
d) 273 K
Explanation: At N.T.P. the temperature is 0 0C. Therefore at absolute condition, the temperature will be, -2730C. The temperature will be equal to 0 K.
7. Which of the following reaction is not involved in combustion calculations?
a) CH4 + 2O2 → CO2 + 2H2O
b) 2C + O2 → 2CO
c) N2 + O2 → 2NO
d) 2H2 + O2 → 2H2O
Explanation: Combustion means conversion of a reactant in the form of CO2 and H2O. The CO obtained in 2C + O2 → 2CO option is again oxidized to form CO2. The calculation of combustion in done by taking number of moles in consideration.
8. Analysis for combustion of solids is always done on weight basis.
Explanation: The particles in solid are very closely packed due to which its volume does not gives accurate values of combustion. To overcome this problem, the combustion for solid is always calculated on the basis of weight. For gases and liquid the analysis is done on volume basis.
9. Find the volume of air required for complete combustion of 1 m3 of acetylene and the weight of air necessary for the combustion of 1 kg of fuel?
a) 10.6 m3 and 12.526 kg
b) 12.1 m3 and 15.5 kg
c) 11.9 m3 and 13.378 kg
d) 12.0 m3 and 14.965 kg
Explanation: The reaction involved in this reaction will be-
2C2H2 + 5O2 → 4CO2 + 2H2O
Therefore air required per m3 of C2H2 = 2.5 × (100/21) = 11.9 m3
Now oxygen supplied per kg of fuel = (162/52)
Therefore, air supplied per kg of C2H2 = (160/52) × (100/23) = 13.378 kg.
10. The % analysis by volume of product gas is H2 – 18.3%, CH2 – 3.4%, CO – 25.4%, CO2 – 5.1% and N2 – 47.8%. Calculate the volume of air required m3 of the gas?
a) 1.3643 m3
b) 1.5646 m3
c) 1.7643 m3
d) 1.9643 m3
Explanation: Total moles of O2 required = 28.65
Therefore, air required for 100 moles of fuel gas = 28.65 × (100/21) = 136.43 mols
Therefore air required per m3 of gas = (136.43/100) = 1.3643 m3.
Sanfoundry Global Education & Learning Series – Engineering Chemistry.
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