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

Question

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

Olivia Anton · Accepted Answer

#C_7H_16# would be the limiting reagent.

Starting from the balanced chemical equation

#C_7H_16 + 11O_2 -> 7CO_2 + 8H_2O#

we can see that we have a #1:11# mole ratio between #C_7H_16# and #O_2#; that is, for every mole of #C_7H_16# used in the reaction, #11# moles of #O_2# are required.

So, the number of #C_7H_16# moles, knowing that its molar mass is #100g/(mol), is

#n_(C_7H_16) = m_(C_7H_16)/(molarmass) = (19.2g)/(100g/(mol)) = 0.2# moles

The number of #O_2# moles is (its molar mass is #32g/(mol)#)

#n_(O_2) = m_(O_2)/(molarmass) = (120g)/(32g/(mol)) = 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_7H_16# is the limiting reagent.

Cooper Adams · Answer

undefined To determine the limiting reagent, calculate the moles of each reactant using their respective molar masses. Then, use stoichiometry to find the amount of product produced by each reactant. The reactant that produces the least amount of product is the limiting reagent.

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

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