What would be the limiting reagent if 19.2 grams of C7H16 were reacted with 120 grams of O2? C7H16+11O2=>7CO2+8H2O

Dec 13, 2014

${C}_{7} {H}_{16}$ would be the limiting reagent.

Starting from the balanced chemical equation

${C}_{7} {H}_{16} + 11 {O}_{2} \to 7 C {O}_{2} + 8 {H}_{2} O$

we can see that we have a $1 : 11$ mole ratio between ${C}_{7} {H}_{16}$ and ${O}_{2}$; that is, for every mole of ${C}_{7} {H}_{16}$ used in the reaction, $11$ moles of ${O}_{2}$ are required.

So, the number of ${C}_{7} {H}_{16}$ moles, knowing that its molar mass is 100g/(mol), is

${n}_{{C}_{7} {H}_{16}} = {m}_{{C}_{7} {H}_{16}} / \left(m o l a r m a s s\right) = \frac{19.2 g}{100 \frac{g}{m o l}} = 0.2$ moles

The number of ${O}_{2}$ moles is (its molar mass is $32 \frac{g}{m o l}$)

${n}_{{O}_{2}} = {m}_{{O}_{2}} / \left(m o l a r m a s s\right) = \frac{120 g}{32 \frac{g}{m o l}} = 3.8$ moles

However, the number of moles needed is

n_(O_2) = n_(C&H_16) * 11 = 0.2 * 11 = 2.2# moles

This means that we have excess ${O}_{2}$ ($3.8 - 2.2 = 1.6$ moles) and ${C}_{7} {H}_{16}$ is the limiting reagent.