When 4.0 mol of CCl4 reacts with an excess of HF, 3.0 mol of CCl2F2 (Freon) is obtained. The equation for the reaction is
CCl4(l) + 2HF(g) --> CCl2F2(l) + 2HCl9g)
State which of the statements are true about the reaction and make the false statements true.
(a) The theoretical yield for CCl2F2 is 3.0 mol.
(b) The theoretical yield for HCl is 71 g.
(c) The percent yield for the reaction is 75%.
(d) The theoretical yield cannot be determined unless the exact amount of HF is given.
(e) From just the information given above, it is impossible to calculate how much HF is unreacted.
(f) For this reaction, as well as for any other reaction, the total number of moles of reactants is equal to the total number of moles of product.
(g) Half a mole of HF is consumed for every mole of CCl4 used.
(h) At the end of the reaction, no CCl4 is theoretically left unreacted.
I'm kind of confused on how to figure out the theoretical yield and percent yield using the information.
c is correct.
1 mole CCl4 produces 1 mole Freon. So 4 moles CCl4 should produce 4.0 moles Freon. 4.0 moles Freon is the theoretical yield. The problem states that 3.0 moles Freon were produced; therefore, the %yield = (3.0/4.0)*100 = 75%.
c is not the only true statement in the list.
Okay thank you!
What is the density of Freon-11 (CFCl3) at 120°C and 1.5 atm?
what will happen when 3.00 grams of CCI4 reacts with 3.00 grams of HF?
what will happen when 3.00 grams of CCI4 reacts with 3.00 grams of HF?
Well, let's clown around and figure this out together!
(a) The theoretical yield for CCl2F2 is 3.0 mol: True! In the balanced equation, the ratio of CCl4 to CCl2F2 is 1:1, so if 4.0 mol of CCl4 reacts, you would expect to get 4.0 mol of CCl2F2. However, since only 3.0 mol of CCl2F2 is obtained, that's the theoretical yield.
(b) The theoretical yield for HCl is 71 g: False! The molar mass of HCl is 36.5 g/mol, and according to the balanced equation, for every mole of CCl4 used, 2 moles of HCl are produced. So, the theoretical yield of HCl would be 4.0 mol x 2 mol x 36.5 g/mol = 292 g, not 71 g.
(c) The percent yield for the reaction is 75%: True! Percent yield is calculated by dividing the actual yield (3.0 mol) by the theoretical yield (4.0 mol) and multiplying by 100. So, (3.0 mol / 4.0 mol) x 100 = 75%.
(d) The theoretical yield cannot be determined unless the exact amount of HF is given: False! Theoretical yield can be calculated based on the stoichiometry of the balanced equation. The molar ratio between CCl4 and CCl2F2 allows us to determine the theoretical yield of CCl2F2 regardless of the exact amount of HF used.
(e) From just the information given above, it is impossible to calculate how much HF is unreacted: True! The information given does not specify the amount of HF used or the amount of HF consumed, so we cannot determine how much HF is unreacted.
(f) For this reaction, as well as for any other reaction, the total number of moles of reactants is equal to the total number of moles of product: False! In this specific reaction, the total number of moles of reactants (CCl4 + 2HF) is 4.0 mol + 8.0 mol = 12.0 mol, whereas the total number of moles of product (CCl2F2 + 2HCl) is only 3.0 mol + 4.0 mol = 7.0 mol.
(g) Half a mole of HF is consumed for every mole of CCl4 used: True! According to the balanced equation, the molar ratio between CCl4 and HF is 1:2, which means that for every mole of CCl4 used, 2 moles of HF are consumed.
(h) At the end of the reaction, no CCl4 is theoretically left unreacted: False! According to the balanced equation, only 3.0 mol of CCl2F2 is obtained, which means that 1.0 mol of CCl4 is left unreacted. So, at the end of the reaction, some CCl4 is theoretically still unreacted.
Hope that clears things up for you! Remember, chemistry can be fun, especially when clowns like me are around to help.
To find the theoretical yield for a reaction, you need to use stoichiometry to determine the mole ratio between the reactants and products. The mole ratio can be obtained from the balanced equation.
Let's start by calculating the theoretical yield of CCl2F2 (Freon).
According to the balanced equation:
1 mol of CCl4 reacts with 2 mol of HF to produce 1 mol of CCl2F2.
Given that 4.0 mol of CCl4 reacts, we can determine the theoretical yield of CCl2F2 by setting up a proportion:
(4.0 mol CCl4) / (1 mol CCl4) = (x mol CCl2F2) / (1 mol CCl2F2)
Solving for x gives us: x = 4.0 mol CCl2F2
Therefore, statement (a) is true. The theoretical yield for CCl2F2 is indeed 4.0 mol, not 3.0 mol. So, to make this statement true, we need to change the theoretical yield value to 4.0 mol.
Now let's move on to statement (b), which states that the theoretical yield of HCl is 71 g. To determine the theoretical yield of HCl, we can use the same approach as before.
According to the balanced equation:
2 mol of HF reacts to produce 2 mol of HCl.
Given that 4.0 mol of CCl4 reacts, we can calculate the moles of HF required:
(4.0 mol CCl4) x (2 mol HF / 1 mol CCl4) = 8.0 mol HF
From this, we know that 8.0 mol of HF will react to produce 8.0 mol of HCl.
To convert moles of HCl to grams, we need to use the molar mass of HCl, which is approximately 36.5 g/mol.
The theoretical yield of HCl can be calculated as follows:
(8.0 mol HCl) x (36.5 g HCl / 1 mol HCl) = 292 g HCl
Therefore, statement (b) is false. The theoretical yield for HCl is 292 g, not 71 g.
Moving on to statement (c) about the percent yield, the percent yield is calculated by dividing the actual yield (in this case, 3.0 mol of CCl2F2) by the theoretical yield (4.0 mol of CCl2F2) and multiplying by 100%.
Percent yield = (3.0 mol / 4.0 mol) x 100% = 75%
So, statement (c) is true. The percent yield for the reaction is indeed 75%.
Statement (d) is false. The theoretical yield can be determined using stoichiometry, as explained above, without needing the exact amount of HF.
Statement (e) is true. With only the given information, it is not possible to calculate how much HF is unreacted. Additional information is required, such as the actual amount of HF used or the actual yield of a byproduct.
Statement (f) is true. In any balanced chemical equation, the total number of moles of reactants is equal to the total number of moles of products. This is a fundamental principle of stoichiometry.
Statement (g) is true. According to the balanced equation, 2 moles of HF are required to react with 1 mole of CCl4. Therefore, half a mole of HF is consumed for every mole of CCl4 used.
Statement (h) is false. At the end of the reaction, there may still be unreacted CCl4 present, as the reaction may not proceed to completion.