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

May 22, 2017

${O}_{2}$

#### Explanation:

If we assume here the products are $C {O}_{2} \left(g\right)$ and ${H}_{2} O \left(g\right)$ (the propane is combusted in air), the equation looks like

${C}_{3} {H}_{8} \left(l\right) + 5 {O}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right)$

To find the limiting reactant, let's convert the given masses of ${C}_{3} {H}_{8}$ 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 ${C}_{3} {H}_{8}$ 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 ${C}_{3} {H}_{8}$, 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.