350.ml of .25M KOH is reacted with 250.ml of .30M H3PO4. Determine which is the limiting reagent, the molarity of the salt formed, and the final molarity of the reactant that was in excess.
350 mL KOH x 0.25M= 87.5 mmoles
250 mL H3PO4 x 0.30M = 75 mmoles.
..........KOH + H3PO4 ==> KH2PO4 + H2O
initial...87.5...75.0......0........0
change...-75.0..-75.0.....+75.....+75
equil.....12.5....0.......75........75
BUT the KH2PO4 (75 mmoles of it) can react with what is left of the KOH.
.......KOH + KH2PO4 ==> K2HPO4 + H2O
init...12.5...75.........0........0
change -12.5..-12.5......12.5.....12.5
equil...0......62.5......12.5....12.5
So KOH is the limiting reagent since it ran out first. The problem is stated as if only one of the H ions on H3PO4 reacts. The salt concn (you pick out the one the question wants because there are two of them) is M = moles/L soln. The molarity of the one in excess is moles/L also. NOte the volume is 250 + 350 mL and convert to L.
To determine the limiting reagent, we need to compare the number of moles of KOH and H3PO4.
First, let's calculate the number of moles of KOH:
Molarity of KOH = 0.25 M
Volume of KOH = 350 mL = 0.35 L
Number of moles of KOH = Molarity × Volume
= 0.25 M × 0.35 L
= 0.0875 moles
Next, let's calculate the number of moles of H3PO4:
Molarity of H3PO4 = 0.30 M
Volume of H3PO4 = 250 mL = 0.25 L
Number of moles of H3PO4 = Molarity × Volume
= 0.30 M × 0.25 L
= 0.075 moles
From the calculations, we can see that the moles of KOH (0.0875 moles) are greater than the moles of H3PO4 (0.075 moles). Therefore, H3PO4 is the limiting reagent.
To determine the molarity of the salt formed, we need to determine the mole ratio between the limiting reagent (H3PO4) and the product (the salt).
The balanced equation for the reaction is:
3 KOH + H3PO4 → K3PO4 + 3 H2O
From the balanced equation, we can see that each mole of H3PO4 reacts with 3 moles of KOH to form 1 mole of the salt (K3PO4).
Therefore, the molarity of the salt formed is equal to the moles of the salt divided by the total volume:
Molarity of the salt = Moles of the salt / Total volume
= (0.075 moles) / (0.35 L + 0.25 L)
= 0.075 moles / 0.6 L
= 0.125 M
Finally, to determine the final molarity of the reactant that was in excess (KOH in this case), we need to find the remaining moles of the excess reactant after the reaction.
The moles of the excess reactant can be calculated by subtracting the moles of the limiting reagent from the initial moles of the excess reactant.
Initial moles of KOH = 0.0875 moles
Moles of KOH used (reacted with H3PO4) = Moles of H3PO4 (0.075 moles) / Stoichiometric coefficient of KOH (3 moles)
= 0.075 moles / 3
= 0.025 moles
Moles of the excess reactant (KOH) = Initial moles of KOH - Moles of KOH used
= 0.0875 moles - 0.025 moles
= 0.0625 moles
Finally, we can calculate the final molarity of the excess reactant:
Final Molarity of the excess reactant = Moles of the excess reactant / Total volume
= (0.0625 moles) / (0.35 L + 0.25 L)
= 0.0625 moles / 0.6 L
= 0.104 M
Therefore, the final molarity of the KOH (the reactant that was in excess) is 0.104 M.
To determine the limiting reagent, we need to calculate how many moles of each reactant are present. The reaction between KOH and H3PO4 is as follows:
3 KOH + H3PO4 → K3PO4 + 3 H2O
First, let's calculate the number of moles of KOH:
Volume of KOH solution = 350 mL = 0.350 L
Molarity of KOH solution = 0.25 M
Using the formula: Moles = Volume x Molarity
Moles of KOH = 0.350 L x 0.25 M = 0.0875 moles
Now, let's calculate the number of moles of H3PO4:
Volume of H3PO4 solution = 250 mL = 0.250 L
Molarity of H3PO4 solution = 0.30 M
Moles of H3PO4 = 0.250 L x 0.30 M = 0.075 moles
According to the balanced chemical equation, the stoichiometric ratio between KOH and H3PO4 is 3:1. This means that 3 moles of KOH react with 1 mole of H3PO4.
Since the stoichiometry ratio is 3:1 and the moles of H3PO4 are less than 1/3 of the moles of KOH, H3PO4 is the limiting reagent.
Now, let's calculate the molarity of the salt formed, K3PO4.
From the balanced chemical equation, we see that the stoichiometry ratio between H3PO4 and K3PO4 is 1:1. Therefore, the moles of K3PO4 formed are equal to the moles of H3PO4.
So, the molarity of K3PO4 = moles of K3PO4 / Volume of K3PO4 solution
Moles of K3PO4 = 0.075 moles
Volume of K3PO4 solution = (Volume of H3PO4 solution + Volume of KOH solution) = (250 mL + 350 mL) = 0.600 L
Molarity of K3PO4 = Moles of K3PO4 / Volume of K3PO4 solution
= 0.075 moles / 0.600 L
= 0.125 M
Finally, let's determine the final molarity of the reactant that was in excess, which in this case is KOH.
Since H3PO4 is limiting, we need to determine how much H3PO4 is required to react with all the KOH. The stoichiometry ratio between H3PO4 and KOH is 1:3.
Moles of H3PO4 required = 3 x Moles of KOH
= 3 x 0.0875 moles
= 0.2625 moles
Now, let's calculate the final molarity of KOH:
Volume of KOH solution = 0.350 L
Moles of KOH reacted = Moles of H3PO4 required = 0.2625 moles
Molarity of KOH (excess) = Moles of KOH reacted / Volume of KOH solution
= 0.2625 moles / 0.350 L
= 0.750 M
Therefore, H3PO4 is the limiting reagent, the molarity of the salt formed (K3PO4) is 0.125 M, and the final molarity of the KOH is 0.750 M.