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!
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.
thank you! :D
Sure, I can help you understand the process and answer your questions step-by-step.
1) To find the theoretical yield of Freon (CCl2F2), we need to calculate the limiting reactant. This is the reactant that will be completely consumed in the reaction and determines the maximum amount of product that can be formed.
To do this, we need to determine the number of moles of each reactant. The molar mass of HF is approximately 20g/mol and the molar mass of CCl4 is approximately 153.82g/mol.
Given that we have 1.0kg of HF and 1.0kg of CCl4, we can convert these masses into moles using their respective molar masses:
Number of moles of HF = (mass of HF)/(molar mass of HF) = (1000g)/(20g/mol) = 50 mol
Number of moles of CCl4 = (mass of CCl4)/(molar mass of CCl4) = (1000g)/(153.82g/mol) = 6.49 mol
From the balanced chemical equation, 2 moles of HF react with 1 mole of CCl4 to produce 1 mole of CCl2F2. Therefore, the ratio of moles is 2:1. Since we have fewer moles of CCl4, it is the limiting reactant.
So, the theoretical yield of Freon (CCl2F2) is determined by the number of moles of CCl4. The molar mass of CCl2F2 is approximately 121g/mol.
Theoretical yield of CCl2F2 = (moles of CCl4) x (molar mass of CCl2F2)
= 6.49 mol x 121g/mol
≈ 785.29 g
Therefore, the theoretical yield of Freon is approximately 785.29g.
2) To calculate the percent yield, we need to compare the actual yield (0.44kg) with the theoretical yield (785.29g) and calculate the percentage.
Percent yield = (actual yield/theoretical yield) x 100
= (0.44kg/785.29g) x 100
≈ 0.056 x 100
≈ 5.6%
Therefore, the percent yield of Freon is approximately 5.6%.
3) From the balanced chemical equation, we can see that 2 moles of HCl are produced for every 2 moles of HF consumed. Since we had 50 moles of HF, the theoretical yield of HCl would be 50 moles as well. The molar mass of HCl is approximately 36.5g/mol.
Theoretical yield of HCl = (moles of HF) x (moles of HCl/moles of HF) x (molar mass of HCl)
= 50 mol x (2 mol HCl/2 mol HF) x (36.5 g/mol)
= 50 mol x 1 x 36.5 g/mol
= 1825 g
Therefore, the theoretical yield of HCl is 1825g.
4) To calculate the total product mass, we need to add up the masses of Freon (CCl2F2) and HCl.
Total product mass = theoretical yield of CCl2F2 + theoretical yield of HCl
= 785.29 g + 1825 g
≈ 2610.29 g
So, the total product mass that should have been produced is approximately 2610.29g.
The total mass of reactants (1kg + 1kg = 2000g) is less than the total product mass. This difference is due to the presence of excess reactant in the reaction. In this case, excess HF is left over after the reaction has gone to completion.
5) The difference between the total mass of reactants and the total product mass is the mass of excess HF that did not react. To find the mass of excess HF, we need to subtract the mass of HF that was consumed in the reaction from the initial mass of HF.
Mass of excess HF = Initial mass of HF - Mass of HF consumed
= 1000g - (moles of HF consumed x molar mass of HF)
= 1000g - (50 mol x 20g/mol)
= 1000g - 1000g
= 0g
Therefore, the difference between the total mass of reactants and the total product mass is zero, meaning there is no excess HF remaining. All the HF was consumed in the reaction.
I hope this helps you understand the process and answers your questions. Let me know if you have any further doubts!
No problem! I'll break it down step by step for you:
1) The theoretical yield of Freon can be calculated by using stoichiometry, which is a way to relate the number of moles of reactants and products in a chemical reaction. First, we need to convert the mass of HF and CCl4 to moles:
1.0 kg HF * (1 mol HF / molar mass of HF) = X moles HF
1.0 kg CCl4 * (1 mol CCl4 / molar mass of CCl4) = Y moles CCl4
Now, we determine the limiting reactant, which is the one that will be completely consumed and limits the amount of product that can be formed. To do this, we compare the mole ratios of the reactants to the balanced equation:
2 mol HF : 1 mol CCl4 (from the balanced equation)
If X/Y > 2/1, then CCl4 is the limiting reactant. If X/Y < 2/1, then HF is the limiting reactant.
Once we know the limiting reactant, we can use the mole ratio from the balanced equation to calculate the moles of CCl2F2 that should have been produced:
Z moles CCl2F2 = Y moles CCl4 * (1 mol CCl2F2 / 1 mol CCl4)
Finally, we can convert the moles of CCl2F2 to grams by multiplying by the molar mass of CCl2F2 to obtain the theoretical yield of Freon.
2) The percent yield is a measure of how efficient the reaction was in producing the desired product. It is calculated by dividing the actual yield (0.44 kg CCl2F2) by the theoretical yield and multiplying by 100:
percent yield = (Actual yield / Theoretical yield) * 100
3) To determine the theoretical yield of HCl, we can use the mole ratio from the balanced equation:
2 mol HF : 2 mol HCl (from the balanced equation)
The moles of HCl can be calculated as:
Theoretical yield of HCl = X moles HF * (2 mol HCl / 2 mol HF)
4) The total product mass should be equal to the theoretical yield of CCl2F2 plus the theoretical yield of HCl. To calculate the total product mass, we simply add the two theoretical yields together.
5) The difference between the total product mass and the total mass of reactants (1 kg + 1 kg) is due to the formation of byproducts or incomplete reaction. In real-world chemical reactions, it's often difficult to achieve 100% yield, so there are losses and side reactions that result in a lower actual yield.
I hope this explanation helps you understand the process better! Let me know if you have any further questions.