Question

Let us write the chemical reaction for oxidation of acetylene:
[
2 C_{2} H_{2}+5 O_{2} rightarrow 4 C O_{2}+2 H_{2} O
]
In this reaction, we can see that, 2 moles of ( mathrm{C}_{2} mathrm{H}_{2} ) reacts with 5 moles of ( mathrm{O}_{2} ) Or, 1 mole of ( mathrm{C}_{2} mathrm{H}_{2} ) reacts with ( 5 / 2 ) moles of ( mathrm{O}_{2} )
Under standard conditions, we can say that:
22.44 L of ( mathrm{C}_{2} mathrm{H}_{2} ) reacts with ( (5 / 2) times 22.4 mathrm{L} ) moles of ( mathrm{O}_{2} )
So, 100 cc of ( C_{2} H_{2} ) will reacts with ( (5 / 2) times 100 ) cc ( = ) 250 cc of ( mathrm{O}_{2} )
But, since air contains ( 20 % ) of oxygen by volume, the amount of air needed to react with 100 cc of ( mathrm{C}_{2} mathrm{H}_{2} ) will be ( =250 times(100 / 20)=1250 mathrm{cc} ) of air.

# Orpio UI MerSITY Air contains 20% O, by volume. How many c.c. of air will be required for oxidation of 100 cc of acetylene (C2H2)? Call tooth (1) 1064 cc (2) 212.8 cc 2C2H2 +502 7400, 42- (3) 500 cc (4) 1250 CC

Solution