1a). If you react 16 mol A with 15 mol B in the reaction 4A + 3B → 2C, what is the limiting reactant?
1b). What is the theoretical yield of product C in mol?
3a.) If Pb(NO3)2 were the limiting reactant in this reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2 KNO3 (aq) and KI were added in excess, what chemicals would be found in the filtrate? What would be ALL the chemicals ,including water if present?
3b.) If you add Pb(NO3)2 (aq) to the chemicals in part a^, would there be any chemical reaction? If so, what is the balanced chemical equation including phases of all chemicals. If not, write "No Reaction"
3c.) If you add KI(aq) to the chemicals listed in part a^, would there be any chemical reaction? If so, what is the balanced chemical eq'n including phases of all chemicals. If not, write "No Reaction"
1a). If you react 16 mol A with 15 mol B in the reaction 4A + 3B → 2C, what is the limiting reactant?
1b). What is the theoretical yield of product C in mol?
**well, your reaction has four times the balanced A, but not 4x of B. So B limits the reacation. You will only get 15/4 * 2 C moles
3a.) If Pb(NO3)2 were the limiting reactant in this reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2 KNO3 (aq) and KI were added in excess, what chemicals would be found in the filtrate? What would be ALL the chemicals ,including water if present?
**No all the Pb(NO3)2 would be used up. However, nitrate ions would be present from the KNO3.
3b.) If you add Pb(NO3)2 (aq) to the chemicals in part a^, would there be any chemical reaction? If so, what is the balanced chemical equation including phases of all chemicals. If not, write "No Reaction"
**Yes, adding more leadIINitrate with the other chems present, will get it going again.
3c.) If you add KI(aq) to the chemicals listed in part a^, would there be any chemical reaction? If so, what is the balanced chemical eq'n including phases of all chemicals. If not, write "No Reaction"
No reaction, as the lead nitrate is gone.
1a) To determine the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.
Given:
Reactant A: 16 mol
Reactant B: 15 mol
From the balanced equation: 4A + 3B → 2C
The stoichiometric coefficient ratio of A to B is 4:3.
To find the limiting reactant, we can use the concept of the limiting reactant's moles being completely consumed by the reaction. For this, we can convert the number of moles of both reactants to the number of moles of C using the stoichiometric ratio.
Moles of C from A: 16 mol * (2 mol C / 4 mol A) = 8 mol C
Moles of C from B: 15 mol * (2 mol C / 3 mol B) = 10 mol C
Since the moles of C from reactant B are larger than the moles of C from reactant A, the limiting reactant is reactant A.
1b) The theoretical yield of product C can be determined based on the limiting reactant. 1 mol of A produces 2 mol of C according to the balanced equation. Therefore, with 16 mol of A as the limiting reactant, the theoretical yield of product C in mol would be:
16 mol A * (2 mol C / 4 mol A) = 8 mol C
3a) Assuming that the reaction proceeds to completion, and Pb(NO3)2 is the limiting reactant, the chemicals found in the filtrate would be KNO3 and water (H2O).
In the given reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Pb(NO3)2 reacts with KI to form the precipitate PbI2(s), while the excess KI remains in the solution as KNO3.
3b) If Pb(NO3)2(aq) is added to the chemicals in the filtrate, there will be no chemical reaction because Pb(NO3)2 does not react with KNO3 or water. Therefore, the balanced chemical equation would be "No Reaction."
3c) If KI(aq) is added to the chemicals listed in the filtrate, a chemical reaction would occur. The balanced chemical equation, including the phases of all chemicals, would be:
2KNO3(aq) + PbI2(s) → 2KNO2(aq) + PbI2(s)
1a) To determine the limiting reactant, you need to compare the mole ratio between the reactants and the coefficients in the balanced equation. In this case, the balanced equation is 4A + 3B → 2C. The mole ratio between A and B is 4:3.
Calculate the amount of product that can be formed from 16 mol of A by using the mole ratio. Since the ratio is 4:3, you can calculate as follows: (16 mol A) * (2 mol C / 4 mol A) = 8 mol C.
Calculate the amount of product that can be formed from 15 mol of B: (15 mol B) * (2 mol C / 3 mol B) = 10 mol C.
Comparing the two results, you can see that the amount of product formed from reactant A is less than the amount formed from reactant B. Therefore, the limiting reactant is A.
1b) To find the theoretical yield of product C in moles, you take the amount of the limiting reactant (16 mol A) and calculate the corresponding amount of product using the mole ratio between A and C: (16 mol A) * (2 mol C / 4 mol A) = 8 mol C.
So, the theoretical yield of product C in moles is 8 mol.
3a) When Pb(NO3)2 is the limiting reactant, it completely reacts to form PbI2(s) and 2KNO3(aq). Since KI is added in excess, it will remain in the filtrate.
Therefore, the chemicals found in the filtrate are 2KNO3(aq) and any excess KI(aq) if present.
If water is present, it will also be present in the filtrate.
3b) When Pb(NO3)2 is added to the chemicals in part 3a, there will be a chemical reaction. The balanced chemical equation, including phases, is as follows:
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
3c) When KI is added to the chemicals listed in part 3a, there will not be a chemical reaction because KI is already in excess. Therefore, the answer is "No Reaction".