Determine the theoretical yield of C when each of the following amounts of A and B are allowed to react in the generic reaction: A+2B-->3C
1) 1mol A; 1 mole B
2) 1mole A; 3 mole B
3) 2mole A; 2 mole B
4) 32mole A; 68 mole B
I will do 1) for you in detail.
A + 2B ==> 3C
mols A = 1. How many moles of C will that form? Using the coefficients in the balanced equation, convert moles A to moles C. 1 mole A x (3 moles C/1 mole A) = 3 moles C.
moles B = 1. How many moles of C will that form? Same process as above, convert moles B to moles C. 1 mole B x (3 moles C/2 moles B) = 3/2 = 1.5 moles C.
3 moles C and 1.5 moles C don't agree and one of those answers must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. So 1.5 moles C is the theoretical yield and B is the limiting reagent. (The question doesn't ask but we can calculate how much A isused up and how much is un-reacted). That will be 1 mole B x (1 mole A/2 moles B) = 1/2 or 0.5 mole A used; therefore, there is 0.5 mole A remaining un-reacted.)
1) Well, according to the reaction, we need 2 moles of B to react with 1 mole of A to produce 3 moles of C. However, since we only have 1 mole of B, it is the limiting reagent. So, the theoretical yield of C would be less than 3 moles. But hey, on the bright side, it's still more than a big fat zero!
2) In this case, we have a nice balance of 1 mole of A and 3 moles of B. So, the mole ratio is maintained and we can expect to produce 3 moles of C. It's like a perfect harmony between A and B, creating a beautiful symphony of C.
3) Ah, a bit of a balance issue here. We have 2 moles of A and 2 moles of B. Well, since the ratio calls for 2 moles of B for every 1 mole of A, it seems like we have equal amounts of A and B. That means we can only produce 2 moles of C. It's like trying to ride a unicycle with two wheels – it just doesn't work as well.
4) Woah, that's a lot of A and B! With 32 moles of A and 68 moles of B, the mole ratio still holds true – 1 mole of A reacts with 2 moles of B to give 3 moles of C. So, we can expect to produce 3 times the number of moles of the limiting reagent, which is B in this case. That would be 136 moles of C. That's a whole lot of C! It's like a clown car full of C, just keeps coming and coming.
To determine the theoretical yield of C for each scenario, we need to compare the stoichiometry of the reaction with the given amounts of A and B.
The molar ratio of A to C is 1:3, and the molar ratio of B to C is 2:3. Let's calculate the theoretical yield for each scenario:
1) 1 mole A; 1 mole B:
The limiting reagent is B since its quantity is the same as A. Therefore, the amount of C formed will be determined by the amount of B. The theoretical yield of C is (1 mole B) x (3 moles C / 2 moles B) = 1.5 moles C.
2) 1 mole A; 3 mole B:
The limiting reagent is A since B is in excess. Therefore, the amount of C formed will be determined by the amount of A. The theoretical yield of C is (1 mole A) x (3 moles C / 1 mole A) = 3 moles C.
3) 2 mole A; 2 mole B:
Both A and B are in a 1:1 ratio, so neither is the limiting reagent. The stoichiometry indicates that 2 moles of A will react with 4 moles of B to produce 6 moles of C. Therefore, the theoretical yield of C is 6 moles.
4) 32 mole A; 68 mole B:
The limiting reagent can be determined by comparing the molar ratios of A and B with the stoichiometry of the reaction. The stoichiometry indicates that 32 moles of A will react with 64 moles of B to produce 96 moles of C. Since 68 moles of B are available, B is the limiting reagent. Therefore, the theoretical yield of C is 96 moles.
To summarize:
1) 1 mole A; 1 mole B: Theoretical yield of C is 1.5 moles.
2) 1 mole A; 3 mole B: Theoretical yield of C is 3 moles.
3) 2 mole A; 2 mole B: Theoretical yield of C is 6 moles.
4) 32 mole A; 68 mole B: Theoretical yield of C is 96 moles.
To determine the theoretical yield of C in each of the given reactions, we need to calculate the limiting reactant and then use stoichiometry to find the amount of C produced.
1) 1 mole A; 1 mole B:
To find the limiting reactant, we compare the number of moles of A and B in the reaction. The stoichiometric ratio between A and B is 1:2. So, if we have 1 mole of A, we would need 2 moles of B. Since we have only 1 mole of B, it is the limiting reactant. This means all of the B will be consumed, and only 0.5 moles of A will be used. According to the stoichiometry of the balanced equation, for each mole of B reacted, 3 moles of C are produced. Therefore, the theoretical yield of C is 3 × 1 = 3 moles.
2) 1 mole A; 3 moles B:
Again, we compare the number of moles of A and B. In this case, we have an excess of B since the stoichiometric ratio is 1:2. So, all of A will be consumed, and only 1 mole of B will be used. For each mole of B reacted, 3 moles of C are produced. Therefore, the theoretical yield of C is 3 × 1 = 3 moles.
3) 2 moles A; 2 moles B:
In this case, we have an equal number of moles for both A and B. However, the stoichiometric ratio between A and B is 1:2. So, all of A will be consumed using 2 moles of B. For each mole of B reacted, 3 moles of C are produced. Therefore, the theoretical yield of C is 3 × 2 = 6 moles.
4) 32 moles A; 68 moles B:
To find the limiting reactant, we compare the number of moles of A and B. The stoichiometric ratio between A and B is 1:2. So, in order to react completely, we need 2 moles of B for every 1 mole of A. Since we already have 68 moles of B, it is in excess, and A is the limiting reactant. This means only 32 moles of A will be consumed. According to the stoichiometry, for each mole of B reacted, 3 moles of C are produced. Therefore, the theoretical yield of C is 3 × 32 = 96 moles.
In summary:
1) 1 mol A; 1 mol B: Theoretical yield of C = 3 moles.
2) 1 mol A; 3 mol B: Theoretical yield of C = 3 moles.
3) 2 mol A; 2 mol B: Theoretical yield of C = 6 moles.
4) 32 mol A; 68 mol B: Theoretical yield of C = 96 moles.