We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. If we have 500. mL of 0.632 M lead(II) nitrate solution and add excess potassium iodide, how much solid lead(II) iodide will form (in grams)?
2 KI(aq) + Pb(NO3)2(aq) �¨ PbI2(s) + 2 KNO3(aq)
Thank you. Here are the steps together with the problem. I suggest you print this and keep it. Most stoichiometry problems are worked this way. Limiting reagent problems are slightly (but not much) different. This problem is in molarity and liters so I will stick to that and at the end will show what to do with other problems with solids & gases.
Step 1. Write and balanced the equation. That is done here.
Step 2. Convert what you have [in this case Pb(NO3)2] into mols. mols = M x L = 0.632 x 0.500 = 0.316
Step3. Using the coefficients in the balanced equation, convert mols of what you have [mols Pb(NO)2] into mols of what you want (in this case you want mols PbI2).
0.316 mols Pb(NO3)2 x [1 mol PbI2/1 mol Pb(NO3)2] = 0.316 x 1/1 = 0.316 mol PbI2.
Step 4. Now convert mols what you wanted to grams.
grams = mols x molar mass = 0.316 x molar mass PbI2 = 0.316 x 461 = 145.676 which rounds to 146 to three significant figures.
a. If your problem starts in grams then step one is mols = grams/atomic mass if an element or grams/molar mass if a compound.
b. If your problem starts with a gas, use PV - nRT to solve n = mols or if the gas is at STP, then convert L to mols by L x (1 mol/22.4L) = mols.
Hope this helps.
If you post again please let us know what you don't understand. We can help you better that way.
Goog luck.
You posted a question earlier that is just another stoichiometry problem. Here it is.http://www.jiskha.com/display.cgi?id=1462776118
What about these problems do you not understand?
DrBob222 I just don't understand which steps to take. I got both of these problems wrong so I do not know what I'm doing wrong. If someone could just wal Drk me through how to solve one step by step I might be able to understand it completely.
DrBob222,
Thank you so much! This should be exactly what I need to get me through these types of problems. If I post another question I will be sure to get specific with what I am confused on. Thank you again!
Please forgive the typos. That should be PV = nRT. I got a - sign in there and it should be an = sign.
Well, it seems like you're trying to bring a little bit of chemistry into the mix! I have to admit, I'm more of a joker than a scientist, but I'll give it a shot.
To answer your question, we'll have to use a bit of stoichiometry. The balanced chemical equation tells us that for every 2 moles of KI, we can form 1 mole of PbI2.
First, let's calculate the number of moles of Pb(NO3)2 in the 500 mL solution:
moles of Pb(NO3)2 = concentration (M) x volume (L)
moles of Pb(NO3)2 = 0.632 M x 0.5 L
moles of Pb(NO3)2 = 0.316 mol
Since the ratio between Pb(NO3)2 and PbI2 is 1:1, we know that 0.316 moles of Pb(NO3)2 will produce 0.316 moles of PbI2.
Now, let's calculate the mass of PbI2 using the molar mass of PbI2:
molar mass of PbI2 = atomic mass of Pb + (2 x atomic mass of I)
molar mass of PbI2 = 207.2 g/mol + (2 x 126.9 g/mol)
molar mass of PbI2 = 459 g/mol
mass of PbI2 = moles x molar mass
mass of PbI2 = 0.316 mol x 459 g/mol
mass of PbI2 = 144.444 g
So, approximately 144.444 grams of solid lead(II) iodide will form.
I hope that answered your question, and I also hope it didn't lead to any chemical reactions in your brain!
To determine how much solid lead(II) iodide will form, we need to calculate the limiting reagent and then use stoichiometry to convert it to grams.
First, let's calculate the number of moles of lead(II) nitrate in the solution:
moles of Pb(NO3)2 = volume (in L) x concentration
moles of Pb(NO3)2 = 0.5 L x 0.632 mol/L
moles of Pb(NO3)2 = 0.316 mol
According to the balanced chemical equation, the stoichiometric ratio between Pb(NO3)2 and PbI2 is 1:1. This means that for every one mole of Pb(NO3)2, we will get one mole of PbI2.
Since potassium iodide is in excess, all of the lead(II) nitrate will react to form lead(II) iodide. Therefore, the number of moles of PbI2 formed will be equal to the number of moles of Pb(NO3)2.
moles of PbI2 = 0.316 mol
Now, we can use the molar mass of PbI2 to convert the moles of PbI2 to grams:
molar mass of PbI2 = atomic mass of Pb + 2 x atomic mass of I
molar mass of PbI2 = 207.2 g/mol + 2 x 126.9 g/mol
molar mass of PbI2 = 461 g/mol
mass of PbI2 = moles of PbI2 x molar mass of PbI2
mass of PbI2 = 0.316 mol x 461 g/mol
mass of PbI2 = 145 g
So, when 500 mL of 0.632 M lead(II) nitrate solution reacts with excess potassium iodide, approximately 145 grams of solid lead(II) iodide will form.