Question

The amount of oxygen at STP that is required for the combustion of 4g of ethylene is ?

Answer 1

Your reaction will be

C2H4 + 3O2 -> 2CO2 + 2H2O

4g of ethylene is 1/7 mole

1 ethylene reacts with 3 O2

1/7 will react with x

X= 3/7 moles

At stp so moles=volume/22.4

Volume= moles× 22.4

9.6 litres

Answer 2

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