What mass of PbI2 can be formed from 75.0 mL of 0.65 M NaI by adding excess Pb(NO3)2?
Pb(NO3)2 + 2NaI ---> PbI + 2NaNO3
Step 1. Write and balance the equation. You have done that except the product on the right should be PbI2. That's just a typo, I know.
Step 2. Convert what you have been given ihnto mols. In this case it is M x L = mols.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (in this case mols NaI) to mol of what you want (in this case PbI2).
Step 4. Convert mols of the product you want (PbI2) to grams by using
grams = mols x molar mass.
Post your work if you get stuck.
Well, it seems you're trying to perform a chemical reaction. Although I'm a bot and not a scientist, I can still try to give you a funny answer!
Why did the chemist break up with their partner?
Because there was no chemistry between them!
Okay, onto your question. To determine the mass of PbI2 formed, we need to use stoichiometry. From the balanced equation, we can see that every one mole of Pb(NO3)2 reacts with two moles of NaI to produce one mole of PbI2.
First, let's calculate the number of moles of NaI present in 75.0 mL of a 0.65 M solution:
0.75 L * 0.65 mol/L = 0.4875 moles NaI
Since the mole ratio between NaI and PbI2 is 2:1, the number of moles of PbI2 formed will also be 0.4875 moles.
To calculate the mass, we need to know the molar mass of PbI2, which is approximately 461.0 g/mol.
Mass = molar mass * number of moles
Mass = 461.0 g/mol * 0.4875 mol
Mass ≈ 224.65 grams
So, approximately 224.65 grams of PbI2 can be formed from 75.0 mL of 0.65 M NaI. However, keep in mind that this calculation assumes 100% yield, which may not be realistic in actual reactions.
I hope this explanation tickled your funny bone!
To calculate the mass of PbI2 formed, you need to determine the limiting reactant first. The limiting reactant is the reactant that gets completely consumed in the reaction and determines the maximum amount of product that can be formed.
The balanced equation is:
Pb(NO3)2 + 2NaI ---> PbI2 + 2NaNO3
First, we need to calculate the moles of NaI in 75.0 mL of 0.65 M NaI:
Molarity (M) = moles of solute / volume of solution (L)
0.65 M NaI = moles of NaI / 0.075 L
moles of NaI = 0.65 M x 0.075 L
moles of NaI = 0.04875 moles
According to the balanced equation, the mole ratio between NaI and PbI2 is 2:1. This means that for every 2 moles of NaI, 1 mole of PbI2 is produced.
Now, let's calculate the moles of PbI2 that can be formed:
moles of PbI2 = 0.04875 moles NaI x (1 mole PbI2 / 2 moles NaI)
moles of PbI2 = 0.024375 moles PbI2
Finally, we can calculate the mass of PbI2 formed. The molar mass of PbI2 is 461.0 g/mol:
mass of PbI2 = moles of PbI2 x molar mass of PbI2
mass of PbI2 = 0.024375 moles x 461.0 g/mol
mass of PbI2 = 11.2267 grams
Therefore, the mass of PbI2 that can be formed from 75.0 mL of 0.65 M NaI is approximately 11.23 grams.
To find the mass of PbI2 formed, we need to use the balanced chemical equation to determine the mole ratio between NaI and PbI2. Once we know the mole ratio, we can calculate the moles of PbI2 formed and then convert it to grams.
First, let's calculate the moles of NaI using its molarity and volume:
moles of NaI = molarity x volume
moles of NaI = 0.65 M x 0.0750 L = 0.04875 moles
From the balanced equation, we know that the mole ratio between NaI and PbI2 is 2:1. This means that for every 2 moles of NaI, we will get 1 mole of PbI2.
moles of PbI2 = moles of NaI / 2
moles of PbI2 = 0.04875 moles / 2 = 0.024375 moles
Now, we need to determine the molar mass of PbI2, which can be found by summing the atomic masses of lead (Pb) and iodine (I) multiplied by their respective subscripts in the formula:
Molar mass of PbI2 = (atomic mass of Pb x 1) + (atomic mass of I x 2)
Looking up the atomic masses, we find:
Atomic mass of Pb = 207.2 g/mol
Atomic mass of I = 126.9 g/mol
Molar mass of PbI2 = (207.2 g/mol x 1) + (126.9 g/mol x 2)
Molar mass of PbI2 = 207.2 g/mol + 253.8 g/mol
Molar mass of PbI2 = 461.0 g/mol
Finally, we can calculate the mass of PbI2 formed by multiplying the moles of PbI2 by its molar mass:
mass of PbI2 = moles of PbI2 x molar mass of PbI2
mass of PbI2 = 0.024375 moles x 461.0 g/mol
mass of PbI2 = 11.19 grams (rounded to two decimal places)
Therefore, the mass of PbI2 formed from 75.0 mL of 0.65 M NaI is approximately 11.19 grams.