What is the molar concentration of the lead nitrate,potassium iodine,lead iodine,and the potassium nitrate after you mix 10.00mL of 0.200M Pb(NO3) and 30.00mL of 0.100M KI together balance the equation KI+Pb(NO3)2 PbI2+KNO3
You need to find and use the arrows.
2KI + Pb(NO3)2 ==> PbI2 + 2KNO3
mols KI = M x L = ?
mols Pb(NO3)2 = M x L = ?
This first part is a limiting reagent (LR) problem.
Convert mols KI to mols PbI2.
Convert mols Pb(NO3)2 to mols PbI2.
The smaller number of mols will be the correct value to use and the reagent producing that value is the LR. Twice that will be the mols KNO3. After you know the LR, that will be zero because all of it will be used.
To find how much of the non LR is left, convert mols LR used to mols non-LR used and subtract from the initial amount.
Then convert all of the mols to M by M = mols/L solution.
To determine the molar concentrations of the lead nitrate (Pb(NO3)2), potassium iodine (KI), lead iodine (PbI2), and potassium nitrate (KNO3) after mixing, we need to apply the principles of stoichiometry and dilution.
First, let's write out the balanced chemical equation:
2KI + Pb(NO3)2 -> PbI2 + 2KNO3
Based on the equation, we can see that 2 moles of KI react with 1 mole of Pb(NO3)2 to produce 1 mole of PbI2 and 2 moles of KNO3.
Step 1: Calculate the moles of Pb(NO3)2 and KI used.
- Pb(NO3)2: 10.00 mL * 0.200 mol/L = 2.00 mmol (millimoles)
- KI: 30.00 mL * 0.100 mol/L = 3.00 mmol
Step 2: Determine the limiting reagent.
The limiting reagent is the reactant that is consumed completely, thereby determining the maximum amount of product formed. To find the limiting reagent, we compare the moles of each reactant to the stoichiometric ratio.
According to the balanced equation, the ratio of Pb(NO3)2 to KI is 1:2. Therefore, each mole of Pb(NO3)2 requires 2 moles of KI for a complete reaction.
Since our ratio is 1:2, we can see that 2 mmol of Pb(NO3)2 would require 4 mmol of KI. As we only have 3 mmol of KI, it is the limiting reagent, which means that Pb(NO3)2 is in excess.
Step 3: Calculate the moles of product formed.
From step 2, we determined that KI is the limiting reagent, meaning it will be completely consumed. One mole of KI yields one mole of PbI2.
Therefore, the moles of PbI2 formed are also equal to 3.00 mmol.
Step 4: Calculate the molar concentrations.
The total volume of the solution after mixing is the sum of the volumes of the Pb(NO3)2 and KI solutions: 10.00 mL + 30.00 mL = 40.00 mL = 0.040 L.
To find the molar concentration, divide the moles by the total volume in liters.
- Pb(NO3)2: 2.00 mmol / 0.040 L = 50.00 mmol/L = 50.00 M (Molar)
- KI: 3.00 mmol / 0.040 L = 75.00 mmol/L = 75.00 M
- PbI2: 3.00 mmol / 0.040 L = 75.00 mmol/L = 75.00 M
- KNO3: The concentration of KNO3 is equivalent to the concentration of the limiting reagent, KI. Therefore, it is also 75.00 M.
Hence, the molar concentrations of Pb(NO3)2, KI, PbI2, and KNO3 after mixing will be 50.00 M, 75.00 M, 75.00 M, and 75.00 M, respectively.