AlF3 is almost insoluble in anhydrous HF but dissolves if KF is present.passage of BF3 through the resulting solution causes AlF
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Solution :
Consider a mixture of methane and hydrogen.
Take the basis as 100 moles of the mixture.
The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.
Moles of methane = 0.75 x 100
Moles of hydrogen = 0.25 x 100
The chemical reactions involved during the reaction are :
The fractional conversion of methane is 90%
Number of moles of methane burned during the reaction is = 0.9 x 75
= 67.5
Moles of methane leaving = initial moles of methane - moles of methane burned
= 75 - 67.5
= 7.5 moles
Fractional conversion of hydrogen is 85%
The number of moles of hydrogen burned during the reaction is = 0.85 x 25
= 21.25
Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned
= 25 - 21.25
= 3.75 moles
Methane undergoing complete combustion is 95%.
formed is = 0.95 x 67.5
= 64.125 moles
formed is = 0.05 x 67.5
= 3.375 moles
Oxygen required for the reaction is as follows :
From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.
Hence, oxygen required is = 2 x 75
= 150 moles
From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.
Hence, oxygen required is = 0.5 x 25
= 12.5 moles
Therefore, total oxygen is = 150 + 12.5 = 162.5 moles
Air is 25% excess.
SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles
Amount of nitrogen =
= 764.136 moles
Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3
Oxygen consumed in reaction 1 :
1 mole of methane requires 2 moles of oxygen for complete combustion
= 2 x 64.125
= 128.25 moles
1 mole of methane requires 1.5 moles of oxygen for partial combustion
= 1.5 x 3.375
= 5.0625 moles
From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen
= 0.5 x 21.25
= 10.625 moles.
Total oxygen consumed = 128.25 + 5.0625 + 10.625
= 143.9375 moles
Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3
Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles
Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles
Amount of steam in reaction 3 = 21.25 moles
Total amount of steam = 128.25 + 6.75 + 21.25
= 156.25 moles
The composition of stack gases are as follows :
Number of moles of carbon dioxide = 64.125 moles
Number of moles of carbon dioxide = 3.375 moles
Number of moles of methane = 7.5 moles
Number of moles of steam = 156.25 moles
Number of moles of nitrogen = 764.136 moles
Number of moles of unused oxygen = 59.1875 moles
Number of moles of unused hydrogen = 3.75 moles
Total number of moles of stack gas
= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75
= 1058.32 moles
Concentration of carbon monoxide in the stack gases is
= 3189 ppm
b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.