What would be the limiting reagent if 19.2 grams of C7H16 were reacted with 120 grams of O2? C7H16+11O2=>7CO2+8H2O Chemistry Stoichiometry Limiting Reagent 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. Answer link Related questions How do you determine how much of the excess reactant is left over? Also, how do you determine... What is the limiting reactant in a Grignard reaction? Question #19e8d Question #c5be6 Question #19e95 Question #dc361 Question #b983d Question #19e99 Question #52b92 Question #3f1d9 See all questions in Limiting Reagent Impact of this question 5867 views around the world You can reuse this answer Creative Commons License