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

1 Answer
Stefan V. · Media Owl
Dec 13, 2014

#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.

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