Fundamentals M.5 Part B

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Chin_Alyssa_3I
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Joined: Fri Jul 22, 2016 3:00 am

Fundamentals M.5 Part B

Postby Chin_Alyssa_3I » Tue Sep 27, 2016 11:50 am

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

Evan Lee 2D
Posts: 18
Joined: Wed Sep 21, 2016 2:58 pm

Re: Fundamentals M.5 Part B

Postby Evan Lee 2D » Tue Sep 27, 2016 3:21 pm

From the given we know that the reaction is carried out with 12 mol ClO2 and 5 mol BrF3. From how I thought of it, I simply doubled the stoichiometric coefficients of the entire reaction. So, when 12 mol Cl02 is used, that means the reaction looks like this:

12 ClO2 + 4 BrF3 ---> 12 ClO2F + 2Br2.

It's the same reaction, just slightly altered because of the given values. You use the mole ratios to determine how many moles of product is used. When 6 moles of ClO2 is used, we know that 6 moles of ClO2F is produced, just by understanding that the two are in a 1:1 mole ratio. In this case, 12 moles of ClO2 is used, meaning that 12 moles of ClO2F is produced. The same goes for the other reactants and products. Both in the given reaction and the rewritten reaction, ClO2 and Br2 remain in a 6:1 ratio. ClO2 and BrF3 remain in a 3:2 ratio.

So to answer the question, we know that 12 moles of the product ClO2F is produced and 2 moles of the product Br2. Now, since only 4 BrF3 was needed for the reaction, but reaction is carried out with 5 moles of BrF3, then that is the excess reactant. So there is one mole of excess of BrF3.

Hope this helped!

Chin_Alyssa_3I
Posts: 34
Joined: Fri Jul 22, 2016 3:00 am

Re: Fundamentals M.5 Part B

Postby Chin_Alyssa_3I » Tue Sep 27, 2016 9:27 pm

Thank you so much!


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