This is a limiting reagent (LR) problem and you solve these by essentially working two simple stoichiometry problems like the Zn/HCl problem below.
Pb(ClO3)2 + 2NaI ==> PbI2(s) + 2NaClO3
mols Pb(ClO3) = M x L = ?
mols NaI = M x L = ?
Now using the coefficients in the balanced equation, convert mols Pb(ClO3)2 to mols PbI2.
Do the same for mols NaI to mols PbI2.
It is likely that these two values for mols PbI2 will not agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now use the smaller value of PbI2 and convert it to grams. g = mols x molar mass.
You see, you just work two of the simple problems [Pb(ClO3)2 to PbI2 and NaI to PbI2], take the smaller value which identifies the LR, then convert that to grams just as before.
Write the balanced equation for (aq) Pb(ClO3)2 and (aq) NaI. Include phases.
What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with .250 L of .110 M NaI? Assume the reaction goes to completion.
6 answers
Are the moles of Pb(ClO3)2 .01375?
I don't know. I assumed you knew M Pb(ClO3)2 from the phrase "concentrated solution Pb(ClO3)2".
It's hard for me to wrap my brain around. I just learned all of this stuff! AH
6.34 g??
Yes, I would go with 6.34. What we are doing is assume a concentrated solution is enough to make it the excess reagent, then work the simple stoichiometric problem.
1 answer