You are given 4.5 moles of O_2O2 to react with 1.60 x10^21.60x102 g C_2H_4C2H4. Upon completion of the reaction, will there be any remaining C_2H_4C2H4?

1 Answer
Nov 16, 2016

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)H2C=CH2(g)+3O2(g)2CO2(g)+2H2O(l)

For complete combustion, 28*g28g of ethylene thus requires 96*g96g of dioxygen.

Here, "moles of ethylene"moles of ethylene == (1.60xx10^2*g)/(28.05*g*mol^-1)=5.70*mol1.60×102g28.05gmol1=5.70mol.

And, this quantity of ethylene requires, 3xx5.70*molxx32.00*g*mol^-13×5.70mol×32.00gmol1 "dioxygen gas"dioxygen gas == 547.6*g547.6g, approx. 17*mol17mol.

However, we can certainly represent incomplete combustion:

H_2C=CH_2(g) +2O_2(g) rarr 2CO(g) + 2H_2O(l)H2C=CH2(g)+2O2(g)2CO(g)+2H2O(l);

And H_2C=CH_2(g) +O_2(g) rarr 2C(s) + 2H_2O(l)H2C=CH2(g)+O2(g)2C(s)+2H2O(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.