Acetylene C2H2 burns in air forming carbon dioxide CO2 and water vapour H2O .Calculate the volume of air required to completely burn 50 cc of acetylene. Assume air contains 20% oxygen...
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Answered by
2
C2H2 + 5/2O2 ----------------> 2CO2 + H2O
Burning of 1 mole of C2H2 requires 2.5 mole O2
That means 2.5 x 22.4 L = 56000 cm3 O2
Let total air required is = M
20% of air = 56000 cm3 O2
20 M/100 = 56000 cm3 O2
M = 56000 x 100/20 = 280000 cm3
Burning of 1 mole of C2H2 requires 2.5 mole O2
That means 2.5 x 22.4 L = 56000 cm3 O2
Let total air required is = M
20% of air = 56000 cm3 O2
20 M/100 = 56000 cm3 O2
M = 56000 x 100/20 = 280000 cm3
Answered by
4
Reaction of burning of acetylene in air forming carbon dioxide:
50cc V
2HC=CH + 3O2 = 4CO2 + 2H2O
2 3
Volume:
V= 50 * 3 /2 = 75 cc of Oxygen required for this burning.
If 75 cc is 20 % , then
x is 100%
x= 375 cc of air.
50cc V
2HC=CH + 3O2 = 4CO2 + 2H2O
2 3
Volume:
V= 50 * 3 /2 = 75 cc of Oxygen required for this burning.
If 75 cc is 20 % , then
x is 100%
x= 375 cc of air.
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