Question #b467f

1 Answer
Nathan L.
Jul 9, 2017

Limiting reagent: #"C"_2"H"_3"Br"_3#

Explanation:

We're asked to find the limiting reactant in a chemical reaction given the masses of reactants.

Let's first write the balanced chemical equation for this reaction (look up what forms if you don't know for sure):

#4"C"_2"H"_3"Br"_3 (l) + 11"O"_2 (g) rarr 8"CO"_2 (g) + 6"H"_2"O" (g) + 6"Br"_2 (l)#

To find the limiting reactant, we convert the given masses of reactants to moles, then divide by the respective coefficient in front of it; the limiting reactant is the quantity whose number is lower:

#76.4cancel("g C"_2"H"_3"Br"_3)((1color(white)(l)"mol C"_2"H"_3"Br"_3)/(266.76cancel("g C"_2"H"_3"Br"_3))) = 0.286/(4color(white)(l)"(coefficient)")#

#= color(red)(0.0716#

#49.1cancel("g O"_2)((1color(white)(l)"mol O"_2)/(32.00cancel("g O"_2))) = 1.53/(11color(white)(l)"(coefficient)")#

#= color(blue)(0.140#

Since the number for #"C"_2"H"_3"Br"_3# is lower, #color(red)("C"_2"H"_3"Br"_3)# #sfcolor(red)("is the limiting reagent"#.

Related questions
See all questions in Limiting Reagent
Impact of this question
1752 views around the world
You can reuse this answer
Creative Commons License