#2C_2H_2(l) + 5O_2(g) -> 4CO_2(g) + 2H_2O(g)# If the acetylene tank contains 37.0 mol of #C_2H_2# and the oxygen tank contains 81.0 mol of #O_2#, what is the limiting reactant for this reaction?

1 Answer
Aug 13, 2018

Answer:

Is not dioxygen the limiting reagent....?

Explanation:

We might repropose the combustion reaction as...

#HC-=CH(g) + 5/2O_2(g) rarr 2CO_2(g) + H_2O(l) + Delta_1#

…(I use the half-integral coefficients, because I find the arithmetic EASIER this way). And so each equiv acetylene requires #5/2*"equiv"# dioxygen gas.

Your conditions propose a #37*mol# quantity of acetylene, the which, by the given stoichiometry require...#37*molxx5/2*mol=92.5*mol# dioxygen gas...

You only gots #81*mol# dioxygen gas...and so clearly dioxygen is the limiting reagent. The reaction MIGHT proceed with incomplete combustion...i.e. carbon is oxidized to elemental carbon or carbon monoxide...for which we could write the HYPOTHETICAL reaction..

#HC-=CH(g) + O_2(g) rarr C(s)+CO(g) + H_2O(l) + Delta_2#

Would #Delta_2>Delta_1#? Why or why not? Enquiring minds want to know...

Note that welders can generally adjust the oxygen supply to their acetylene torches, and get reducing or oxidizing flames...