This set of Engineering Chemistry Puzzles focuses on “Numerical Problems Based on Combustion and Fuel Gas Analysis”.
1. How many moles of sulphur are present in 100gm of sulphur di oxide? If the reaction is as follows.
S + O2 → SO2
a) 2
b) 3.6725
c) 6.25
d) 1.5625
View Answer
Explanation: 64 gm of sulphur di oxide is obtained from 32 gm of sulphur,
Therefore for 1 gm of sulphur di oxide = (32/64)*1 = 0.5 gm sulphur
Now for 100gm of sulphur di oxide = 0.5 * 100 = 50 gm sulphur
Now number of moles of sulphur = (mass of sulphur)/ (molecular weight of sulphur)
= 1.5625 moles.
2. What is density of air at normal temperature pressure condition?
a) 998.2 kg/m3
b) 1.290 kg/m3
c) 1290 kg/m3
d) 9.982 kg/m3
View Answer
Explanation: The density of air at normal temperature pressure is 1.290 kg/m3 or 1.290 gm/cm3. The density of water at normal temperature at N.T.P is 998.2 kg/m3.
3. What is the mean molecular weight of air?
a) 28.95
b) 26.4
c) 30.98
d) 24.65
View Answer
Explanation: The mean molecular weight of air can be calculated by taking the mean of the total mass of all the gases present in air, i.e., nitrogen, oxygen and argon. Considering 21 moles of oxygen, 78.06 moles of nitrogen and 0.94 moles of argon, weight of air can be calculated.
4. What is the formulae to calculate the minimum amount of oxygen required for combustion?
a) Minimum O2 required = theoretical O2 required + O2 present in fuel
b) Minimum O2 required = O2 present in fuel
c) Minimum O2 required = theoretical O2 required – O2 present in fuel
d) Minimum O2 required = O2 present in fuel – theoretical O2 required
View Answer
Explanation: The minimum oxygen required should be calculated on the basis, that complete combustion is taking place according to theoretical and stoichiometric combustion reactions. In case of partial combustion, the combustion products are CO.
5. Which product is obtained after irregular combustion of a substance?
a) O2
b) CO
c) CH4
d) CO and O2
View Answer
Explanation: In case of irregular combustion, the combustion products contain both CO and O2. In case of complete combustion O2 is produced and in case of partial combustion CO is produced.
6. How many moles of CO2 are present in the following reaction if 75 kg of carbon is used?
C + O2 + N2 → CO2 + N2
a) 6.25
b) 7.25
c) 5.25
d) 8.25
View Answer
Explanation: 12 gm of carbon is required to produce 48 gm of carbon di oxide,
Therefore for 1 gm of carbon = (48/12)*1 = 4 gm carbon di oxide
Now for 100gm of carbon di oxide = 4* 100 = 400 gm carbon di oxide
Now number of moles of carbon di oxide = (mass of carbon di oxide)/(molecular weight of carbon di oxide) = 6.25 moles.
7. Which part of hydrogen does not under goes combustion?
a) H2
b) OH–
c) H2O
d) H2
View Answer
Explanation: Hydrogen present in the form of moisture does not undergo combustion that is why the combined hydrogen present in coal does not undergo combustion. It is only available hydrogen which is equivalent to (H-O/8) that takes part in combustion.
8. What is the formulae used for calculating the theoretical amount of weight of air required for combustion?
a) Air required = 100/21[(32/12) × C + 8 × (H – O/8) + S] kg
b) Air required = 100/23[(32/12) × C + 8 × (H – O/8) + S] kg
c) Air required = 100/23[(32/12) × C – 8 × (H – O/8) + S] kg
d) Air required = 100/21[(32/12) × C – 8 × (H – O/8) + S] kg
View Answer
Explanation: For complete combustion of 1 kg of solid or liquid fuel, the theoretical amount of air required is
Air required = 100/23[(32/12) × C + 8 × (H – O/8) + S] kg
For volume the percentile is taken as 100/21.
9. How much percentile of excess air is used, if the theoretical amount of air is 95 kg and the actual amount of air present is 100 kg?
a) 5.26
b) 5
c) 10.26
d) 10
View Answer
Explanation: The formulae for calculating the percentile excess amount of air is,
% of excess air = [(actual air used – theoretical air used)/theoretical air used] × 100
= [(100-95)/95)] × 100
= 5.26 %.
10. The mol method is considered to be less convenient and difficult than the stoichiometric weight relationship for calculating the combustion of carbon.
a) True
b) False
View Answer
Explanation: The mol method is more convenient since the number of moles are easy to calculate then the weight of a molecule. Also, weight depends of certain factors which sometimes gives wrong values.
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