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?
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)
5 answers
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.
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.
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.
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!
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.