Asked by Ruffles
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
Answers
Answered by
DrBob222
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.
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.
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