Limiting Reactants

Question

For this reaction, 35.1 g bromine reacts with 13.6 g chlorine gas.

bromine (g) + chlorine (g)  bromine monochloride (g)

What is the maximum mass of bromine monochloride that can be formed?  

What is the FORMULA for the limiting reagent? 

What mass of the excess reagent remains after the reaction is complete?  

Answer 1

For this problem, you must first determine the proper formulas for the various components listed in the problem. Bromine and chlorine are both naturally diatomic elements. So the unbalanced equation is:

Br2 + Cl2 ---> BrCl balanced we have: Br2 + Cl2 ---> 2BrCl

Since masses were given for both reactants, this is a limiting reactant problem...you must first determine what reactant "runs out" first, which would then be the limiting reactant and completion of the calculation from that reactant would provide the theoretical yield. There are two ways to determine the theoretical yield. You can compare one reactant to the other in a stoichiometric calculation or you can simply perform the stoichiometric calculation to determine the amount of mass of BrCl that can be formed from each reactant.

from Br2: 35.1 g Br2 * 1 mol Br2 * 2 mol BrCl * 115.4 g BrCl = 50.7 g BrCl

159.8 g Br2 1 mol Br2 1 mol BrCl

from Cl2 13.6 g Cl2 * 1 mol Cl2 * 2 mol BrCl * 115.4 g BrCl = 44.3 g BrCl

70.90 g Cl2 1 mol Cl2 1 mol BrCl

For answer 1....the maximum amount of BrCl formed is 44.3 g BrCl

For answer 2....the formula for the limiting reactant is Cl2

For the excess reactant, we determine how much Br2 was consumed during reaction with the limiting reactant Cl2 and then subtract that value from the original mass.

13.6 g Cl2 * 1 mol Cl2 * 1 mol Br2 * 159.8 g Br2 = 30.7 g Br2 consumed

70.90 g Cl2 1 mol Cl2 1 mol Br2

excess Br2 = 35.1 g (starting mass) - 30.7 g (consumed) = 4.4 g (excess) Br2

Hope this helps

Answer 2

First, write the correctly balanced equation:

Br₂(g) + Cl₂(g) ===> 2BrCl(g)

To find the maximum mass of BrCl that can form, find the limiting reactant. That will dictate the mass of product that can form. There are several ways to find the limiting reactant. One way is to determine moles of BrCl produced from each:

For Br₂: 35.1 g Br₂ x 1 mole/160 g x 2 moles BrCl/1 mol Br₂ = 0.439 moles BrCl formed

For Cl2: 13.6 g Cl₂ x 1 mole/70 g x 2 mole BrCl/1 mol Cl₂ = 0.389 moles BrCl formed

Thus, Cl₂ is limiting.

Maximum mass BrCl that can be formed = 0.389 mole BrCl x 115 g/mole = 44.7 g

To find mass of excess reactant (Br₂):

initial moles Br₂ = 35.1g/160g = 0.219 moles Br₂

moles used in reacting with Cl₂ = 0.194 (moles Cl₂ used)

moles Br₂ remaining = 0.219 -0.194 = 0.025

mass Br₂ = 0.025 mol x 160 g/mol = 4 g