Question M.5 (Sixth Edition)

[Steve Magana 2I]

Question: Solve this exercise without using a calculator. The reaction 6 ClO2 + 2 BrF3 ----> 6 ClO2F + Br2 is carried out with 12 mol ClO2 and 5 mol BrF3. (a) Identify the excess reactant. (b) Estimate how many moles of each product will be produced and how many moles of the excess reactant will remain.

The part that i have the most trouble understanding is what to do with the 12 mol ClO2 and 5 mol BrF3. Thank you!

[Diane Kang 2F]

(a) To identify the excess reactant use the reaction mole ratio of ClO2 and BrF3, which is 6 mol ClO2 : 2 mol BrF3
(5 mol BrF3) (6 mol ClO2 / 2 mol BrF3) = 15 mol ClO2 , which shows that 5 mol of BrF3 must react with 15 mol ClO2 for BrF3 to go to completion. But since only 12 mol of ClO2 is available, ClO2 is the limiting reactant, making BrF3 the excess reactant.

(b) From (a), you were able to figure out that ClO2 is the limiting reactant, so your estimation of the product quantities should be based on ratios with the products and ClO2.
For Br2: (12 mol ClO2) (1 mol Br2 / 6 mol ClO2) = 2 mol Br2
For ClO2F: (12 mol ClO2) ( 1 mol ClO2F / 1 mol Clo2) = 12 mol ClO2F
For excess reactant BrF3: (12 mol ClO2) (2 mol BrF3 / 6 mol ClO2) = 4 mol BrF3
Only 4 moles of BrF3 react with 12 moles of ClO2. 5 mol BrF3 - 4 mol BrF3 = 1 mol BrF3 remaining.

[Krisdylle Repollo 4H]

The problem already gives you the amount of moles of each reactant (12 mol ClO2 and 5 mol BrF3). To determine the excess reactant, you can use one of moles of these reactants and multiply it by the stoichiometric coefficients in the reaction. For example:

12 mol ClO2 x (2 mol BrF3/6 mol ClO2) = 4 mol BrF3

This shows that you only need 4 mol BrF3 for the reaction, but the problem gives you 5 mol of BrF3. Therefore BrF3 is the excess reactant and ClO2 is the limiting.

(I'm not sure how to do part B though.)