The answer is #0.667#.
The combustion of acetylene can be written as
#2C_2H_2 + 5O_2 -> 4CO_2 + 2H_2O#
The balanced chemical reaction shows a #2:5# mole ratio for #C_2H_2# and #O_2#, a #1:2# ratio for #C_2H_2# and #CO_2#, and a #1:1# ratio for #C_2H_2# and #H_2O#.
Now, for simplicity, let us assume that we have #1# mol of #C_2H_2# to start with ( the equivalence of #26# grams ); given the fact that the oxygen is in excess, acetylene will always act as a limiting reagent.
So, one would get #2# moles of #CO_2# and #1# mole of #H_2O# in the products, for a total of # n_T = 2 + 1 = 3# moles of product.
Therefore, the mole fraction of #CO_2# would be
# n_(CO_2) = (n_(CO_2)/n_T) = 2/3 = 0.667#
In this case, #20%# excess #O_2# would mean that you started with #3# moles of #O_2# instead of #2.5#, but I don't see how this would be relevant other than determining that #O_2# is not a limiting reagent.
