Given the chemical equation, Cl2 + F2 2ClF, how many moles of ClF are produced in the reaction, if 1.50 moles Cl2 are made to react with 1.75 moles F2 ?
I got the answer 3.50 moles but it was wrong. Can anyone help me please.
Cl2 + F2 -> 2ClF
You only have 1.50 mol of Cl2 so you are only going to get 3.00 mol of ClF
You have 0.25 mol of excess F2 left over.
Sorry but I still don't understand your explaining. Can you explain more briefly please. Thanks.
This is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants.
Your 1.50 mols Cl2 will produce 2*1.50 = 3.00 mols ClF if all of the Cl2 reacts and you have an excess of F2.
Your 1.75 mols F2 will produce 2*1.75 = 3.50 mols ClF if you use all of the F2 and have an excess of Cl2. But you don't have an excess of Cl2. You have ONLY 1.50 mols so the yield is determined by the one that will produce the LEAST product. That is Cl2. You can get 3.00 mols ClF and you will have some F2 left over as Damon suggested.
Thanks both of your replied. I got it now :)
To find the number of moles of ClF produced in the reaction, you need to use the concept of mole ratios.
First, let's write down the balanced chemical equation:
Cl2 + F2 -> 2ClF
From the balanced equation, you can see that one mole of Cl2 produces 2 moles of ClF. Therefore, the mole ratio is 1:2 between Cl2 and ClF.
Now, let's calculate the number of moles of ClF produced using the given information:
1. Start with the given moles of Cl2 (1.50 moles) and the mole ratio of Cl2 to ClF (1 mole Cl2: 2 moles ClF).
2. Set up a proportion to find the number of moles of ClF:
(1.50 moles Cl2) = (x moles ClF) / (2 moles ClF)
3. Solve the proportion for x, the number of moles of ClF:
x = (1.50 moles Cl2) * (2 moles ClF / 1 mole Cl2) = 3.00 moles ClF
Therefore, the correct answer is 3.00 moles of ClF, not 3.50 moles.