If 28 g of C_3H_8 and 45 g of O_2 are reacted together, which one would be the limiting reactant?

1 Answer
May 22, 2017

O_2

Explanation:

If we assume here the products are CO_2(g) and H_2O(g) (the propane is combusted in air), the equation looks like

C_3H_8 (l) + 5O_2(g) rightleftharpoons 3CO_2(g) + 4H_2O(g)

To find the limiting reactant, let's convert the given masses of C_3H_8 and O_2 to moles using their molar masses, and then divide the calculated number by the coefficient in front of it in the equation. Whichever number in the end is lower, that reactant is limiting.

Using the molar masses of C_3H_8 and O_2:

28cancel(gC_3H_8)((1molC_3H_8)/(44.11cancel(gC_3H_8))) = color(red)(0.63 mol C_3H_8

45cancel(gO_2)((1molO_2)/(32.00cancel(gO_2))) = (1.4molO_2)/(5"(coefficient)") = color(blue)(0.28 mol O_2

Since O_2 is present in a lower relative amount than C_3H_8, oxygen is the limiting reactant. This problem is a good example of the fact that just because a reactant is present with a lower mass, that doesn't mean it's always the limiting reactant.