Question

10.0 ml mixture of methane and
ethylene was exploded with 30ml
(excess) of oxygen. After cooling
the volume was 21.0 ml. On
treatement with KOH solution the
volume was reduced to 7.0 ml. The
amounts of methane and ethylene
in the mixture respectively are​

Answer 1

Given:

Volume of the mixture = 10ml

Volume of excess of oxygen = 30 ml

Volume of the mixture after cooling = 21 ml

Volume after treatment with KOH = 7 ml

To Find :

The amount of methane and ethylene in the mixture

Solution :

• Let the amount of methane (CH₄) = x ml

• The amount of ethylene(C₂H₄) = (10-x) ml

• The following reactions took place during the process

             $CH_{4}(g)+2O_{2}(g) \rightarrow CO_{2}(g) +2H_{2}O(l)\\ CH_{2} =CH_{2}(g) +3O_{2}(g) \rightarrow 2CO_{2} (g)+H_{2}O(g)$

• The volume of CO₂ absorbed by KOH = 21 - 7 = 14 ml

             $x+2(10-x)=14$

             $x=6ml$

• Volume of CH₄ = 6 ml

       Volume of C₂H₄ = (10-6) = 4 ml

The amounts of methane and ethylene are 6ml, 4ml respectively.