In chemistry, one works with moles and equations. Convert everything in grams to moles and use the equations to convert from one kind of molecule to another.
1. Write the equation and balance it. You have that.
2a. Convert 1 kg HF to moles. moles = g/molar mass.
moles HF = 1,000 g/20.00 =50.00864.1 grams. moles HF
2b. Convert 1,000 g CCl4 to moles. 1,000 g/153.8 = 6.502
In the next step, we use the coefficients in the balanced equation to convert from the moles we have to the moles we want. Note that the fraction we use to convert makes the unit we don't want cancel and leaves the unit to which we wish to convert.)
3a. Using the coefficients in the balanced equation, convert moles HF to moles CCl2F2.
50.00 moles HF x (1 mole CCl2F2/2 mol HF)= 50.00 x (1/2) = 25.00 moles CCl2F2.
3b. Same process, convert moles CCl4 to moles CCl2F2.
6.502 moles CCl4 x (1 mole CCl2F2/1 mole CCl4) = 6.502 moles CCl2F2.
3c. Note that the answers for CCl2F2 in steps 3a and 3b do not agree which means one of them is wrong. The correct one, in limiting regent problems (that's what this is) is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. So the smaller value is 6.502 moles CCl2F2 which makes CCl4 the limiting reagent.
4. Now convert moles product to grams. g = moles x molar mass
g CCl2F2 = 6.502 x 120.91 = 786.156 which I would round to 786.2 grams CCl2F2. This is the theoretical yield of CCl2F2.
Most limiting reagent problems stop here. This problem is about 5 rolled into 1; therefore, don't let the length of this one get to you because it truly is a loooong one.
%yield = (actual yield/theoretical yield)*100 = ??
actual yield is the 440 g CCl2F2 in the problem. Theoretical yield is the 786.2 g above. (440/786.2)*100 = 55.97%
For the theoretical yield of HCl, we need not go through all that stuff above. We already know the limiting reagent is CCl4; therefore, we need only convert moles CCl4 to moles HCl.
6.502 moles CCl4 x (2 moles HCl/1 mole CCl4) = 6.502 x 2 = 13.00 moles HCl.
Now convert to grams (same as steps above); 13.00 moles HCl x molar mass HCl = 13.00 x 36.56 = 473.98 grams HCl which I will round to 474.0 grams.
Total mass that should have been obtained is the theoretical yield of CCl2F2 (786.2 g) + theoretical yield of HCl (474 g) = 1260.2 g
mass on the left must add to mass on the right; however, in this case (a limiting reagent problem) not all of the HF reacted (some was in excess), in fact an excess of 2 kg-1.2602 kg =??
Confirm all of the above yourself. I notice as I've gone though this a couple of times and I've found mistakes. There may be others I didn't see.
I hope this helps you navigate through chemistry. Copy this and use it as a guide for other problems that are limiting reagent problems.
Manufacturing of Freon (CCl2F2):
2 HF(g) + CCl4(I) --> CCl2F2(g) + 2 HCl(g)
Supplies on hand: 1.0kg HF and 1.0 kg CCl4. After the reaction was conducted, 0.44kg CCl2F2 was measured in the lab.
1) What is the theoretical yield of Freon?
2) What was the percent yield?
3) What mass of HCl should have been produced (i.e., theoretical yield of HCl)?
4) How much total product mass should have been produced?
Why is this different from the total mass of reactants (1kg + 1kg) and by how much?
5) What makes up this difference?
I am really confused about all this chemistry ; I can't seem to figure it out. If anyone could help me understand this process, I'd really appreciate it!
2 answers
thank you! :D