You are given 4.5 moles of #O_2# to react with #1.60 x10^2# g #C_2H_4#. Upon completion of the reaction, will there be any remaining #C_2H_4#?

1 Answer
Nov 16, 2016

Answer:

There is insufficient dioxygen to combust the ethylene completely.

Explanation:

We represent the combustion of ethylene by this reaction:

#H_2C=CH_2(g) +3O_2(g) rarr 2CO_2(g) + 2H_2O(l)#

For complete combustion, #28*g# of ethylene thus requires #96*g# of dioxygen.

Here, #"moles of ethylene"# #=# #(1.60xx10^2*g)/(28.05*g*mol^-1)=5.70*mol#.

And, this quantity of ethylene requires, #3xx5.70*molxx32.00*g*mol^-1# #"dioxygen gas"# #=# #547.6*g#, approx. #17*mol#.

However, we can certainly represent incomplete combustion:

#H_2C=CH_2(g) +2O_2(g) rarr 2CO(g) + 2H_2O(l)#;

And #H_2C=CH_2(g) +O_2(g) rarr 2C(s) + 2H_2O(l)#.

Both of these reactions would occur to some extent in any combustion reaction. Given the limited quantity of oxidant, some ethylene would remain unreacted.