M5

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venning-1J
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Joined: Fri Sep 29, 2017 7:04 am

M5

Postby venning-1J » Thu Oct 05, 2017 3:44 pm

I dont understand why in the final reaction, only 2 mol of Br2 are produced. Can some one explain?

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.

Nisarg Shah 1C
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

Re: M5

Postby Nisarg Shah 1C » Thu Oct 05, 2017 4:32 pm

The stoichiometric constants in a balanced chemical equation can be used to illustrate how much of each reactant you need to get a certain amount of product. I'm assuming that by getting part a, you know that ClO2 is the limiting reactant. Knowing that, you can use dimensional analysis to determine how many moles of product will be formed: 12mol ClO2 x (1mol Br2/6mol ClO2) = 2 moles of Br2. That's because for every six moles of ClO2 in the reaction, one mole of Br2 will be produced.

Shawn Patel 1I
Posts: 54
Joined: Thu Jul 27, 2017 3:01 am

Re: M5

Postby Shawn Patel 1I » Thu Oct 05, 2017 9:43 pm

Hey,
In part A, we found that the limiting reactant was ClO2. This means that the amount of product that is created is based on the amount of moles of ClO2 present. We have 12 moles of ClO2, and the ratio of ClO2 to Br2 is 6:1. So that means for every 6 moles of ClO2 there is, 1 mole of Br2 can be made. Since we 12 moles of ClO2, 2 moles of Br2 can be made.

I hope this helps.


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