The average molecular weight of a flue gas having the composition by volume as CO2 = 25%, O2 = 25%, N2 = 50% will be (A) 27.6 (B) 23 (C) 47.3 (D) 42.9

The average molecular weight of a flue gas having the composition by
volume as CO2 = 25%, O2 = 25%, N2 = 50% will be
(A) 27.6
(B) 23
(C) 47.3
(D) 42.9
by

1 Answer

Answer :

(B) 23
by

Related questions

Volumetric composition of flue gas analysed with the Orsat apparatus is : CO2 = 12%, O2 = 8%, CO = nil, N2 = 80%. This flue gas composition indicates that (A) Pure oxygen has been used for combustion (B) Nitrogen percentage in the fuel is very high (C) Excess air has been used for combustion (D) Hydrogen is not present in the fuel

Description : Volumetric composition of flue gas analysed with the Orsat apparatus is : CO2 = 12%, O2 = 8%, CO = nil, N2 = 80%. This flue gas composition indicates that (A) Pure oxygen has been used for combustion ... very high (C) Excess air has been used for combustion (D) Hydrogen is not present in the fuel

Last Answer : (C) Excess air has been used for combustion

A sample of natural gas containing 80% methane (CH4 ) and rest nitrogen (N2 ) is burnt with 20% excess air. With 80% of the combustibles producing CO2 and the remainder going to CO, the Orsat analysis in volume percent is (A) CO2 : 6.26, CO : 1.56, O2 : 3.91, H2O :15.66, N2 : 72.60 (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02 (C) CO2 : 6.39, CO : 1.60, O2 : 3.99, H2O:25.96, N2 :72.06 (D) CO2 : 7.60, CO : 1.90, O2 : 4.75, N2 : 85.74

Description : A sample of natural gas containing 80% methane (CH4 ) and rest nitrogen (N2 ) is burnt with 20% excess air. With 80% of the combustibles producing CO2 and the remainder going to CO, the Orsat analysis in volume percent is (A) CO2 : 6 ... 96, N2 :72.06 (D) CO2 : 7.60, CO : 1.90, O2 : 4.75, N2 : 85.74

Last Answer : (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

The calorific value of a gas of composition CO2 = 20%, H2 = 50%, CH4 = 30%, is y. If the composition of gas is changed to H2 = 50%, CH4 = 30%, CO2 = 10%, N2 = 10%, then the calorific value will be (A) y (B) 0.95 y (C) 1.05 y (D) 1.8 y

Description : The calorific value of a gas of composition CO2 = 20%, H2 = 50%, CH4 = 30%, is y. If the composition of gas is changed to H2 = 50%, CH4 = 30%, CO2 = 10%, N2 = 10%, then the calorific value will be (A) y (B) 0.95 y (C) 1.05 y (D) 1.8 y

Last Answer : (A) y

Air contain 21% O2 and 79% N2 by volume. Calculate the composition in terms percentage by weight.

Description : Air contain 21% O2 and 79% N2 by volume. Calculate the composition in terms percentage by weight.

Last Answer : Basis: 100 kmole air Kmoles of O2 =21 Kmoles of N2 =79 Weight of O2 =21x32=672 kg Weight of N2 =79x28=2212 kg Total weight =2884 kg Weight% of O2 = (weight of O2/ total weight)x100  = (672/2884)x100  = 23.3% Weight% of N2 =(weight of N2/ total weight)x100  = (2212/2884)x100  = 76.7%

The sequence of absorption in flue gas analysis by Orsat's apparatus is respectively (A) CO2 , O2 , CO (B) CO, O2 , CO2 (C) CO2 , CO, O2

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Last Answer : (A) CO2 , O2 , CO

When incomplete combustion loss is high, the flue gas analysis shows large amount of (A) CO2 (B) CO (C) O2 (D) C

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Last Answer : (B) CO

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Last Answer : (B) CO, H2 & CH4

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Last Answer : CO, N2 and H2.

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Last Answer : b) O2

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Last Answer : Combustion analyser

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Last Answer : (a) 8 times

In practical operation of any furnace, zero oxygen percentage or theoretical CO2 percentage in flue gas is rarely achieved, because of (A) Use of non-preheated combustion air (B) Use of pulverised solid fuels (C) Imperfect mixing of fuel & air and infiltration of air (D) Use of excessive positive draft in the furnace

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Last Answer : (A) More of excess air

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Description : A carbonaceous fuel (containing no H2 or hydrocarbons) is burnt and the resulting flue gas contains 21% CO2 . It means that (A) 21% excess air has been used for combustion (B) 21% excess oxygen has ... (C) Complete combustion of fuel has taken place (D) No excess air has been used for combustion

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Last Answer : (D) All (A), (B) and (C)

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Last Answer : (A) CH4 & CO2

Largest constituent of blast furnace gas is (A) N2 (B) CO (C) CO2 (D) H2

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Producer gas comprises mainly of (A) CO & N2 (B) CO & H2 (C) CO2 & N2 (D) CO2 & H2

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Last Answer : (B) CH4

The principal gas evolved from sludge digestion tank is (A) CO(B) CO2 (C) CH4 (D) N2

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Last Answer : (A) CO, CO2 N2 , H2

Synthesis gas is a mixture of (A) CO and H2 (B) N2 and H2 (C) H2 , CH4 and CO (D) CO2 and H2

Description : Synthesis gas is a mixture of (A) CO and H2 (B) N2 and H2 (C) H2 , CH4 and CO (D) CO2 and H2

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A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Description : A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Last Answer : 42.41 ft^3

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