Problem M.5

Moderators: Chem_Mod, Chem_Admin

Posts: 104
Joined: Wed Sep 18, 2019 12:19 am

Problem M.5

Postby AKhanna_3H » Wed Oct 09, 2019 3:11 pm

Could someone please explain how to approach problem M.5? Thanks!

Vanessa Chuang 4F
Posts: 51
Joined: Sat Aug 24, 2019 12:18 am

Re: Problem M.5

Postby Vanessa Chuang 4F » Wed Oct 09, 2019 3:50 pm

M.5 asks the following:
"Solve this exercise without using a calculator. The reaction 6 ClO2(g) + 2 BrF3(l) --> 6 ClO2F(s) + Br2(l) is carried out with 12 mol ClO 2 and 5 mol BrF 3. (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."

First we need to compare the the expected amount of mols of ClO2 used to the amount of mol of BrF3 used. Using the balanced equation, we recognize that for every 1 mol of ClO2 used, we use 2/6 (or 1/3) mol of BrF3. If we compare that ratio against the given amount of ClO2 we are given (12 mol), we expect to use 4 moles of BrF3 (1:1/3 --> 12:4). This means that BrF3 is in excess.
Using the molar ratios, if we use 12 mol of ClO2, we should expect to produce the exact same amount of ClOF (12 mols) and 1/6 the amount of Br2 (2 mols). We will have 1 mol of BrF3 in excess (5 mol avaliable-4 mol used).

Hope that helps!

Posts: 62
Joined: Wed Sep 18, 2019 12:19 am

Re: Problem M.5

Postby Ananta3G » Wed Oct 09, 2019 4:44 pm

This question is just testing if you understand that a) the stoichiometric coefficients before each product or reactant in the balanced equation tells you molar ratios and b) the limiting reactant controls how much product is formed. First, if there are 12 mol of of CIO2 and the reactants are in a 6:2 ratio, you can conclude that you need 12/3 moles of BrF3 which means you only need 4. However, we are given 5 which means BrF3 is in excess. For the second part, you can easily figure out the moles of the excess reactant by doing 5-4 mol which gives 1 mol of BrF3. In order to deduce how much product is made, using the molar ratio of 1 ClO2 to 1 ClO2F, you can get 12 mole of the product and 2 moles of Br2 for the 6:1 ratio of ClO2 to Br2. Hope my answers helps clear up the concepts a bit! :)

Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest