[potatoeater567]

Hi all, I want to show you the work for this problem I did, I want to know if I did it right. If I didn't do it right, could someone explain it to me, please?

Problem: "How many mL of 0.100 M KCl(aq) are required to react completely with 4.325 grams of solid Pb(NO3)2 to form PbCl2?"

My work:

I first balanced the equation and got 2KCl(aq) + Pb(NO3)2(s) = 2KNO3(aq) + PbCl2(s)

Then: (4.325 g. of Pb(NO3)2)(1 mol Pb(NO3)2/331.2098 g. Pb(NO3)2)(2 mol KCl/1 mol Pb(NO3)2)= 0.02612 mol KCl


Now that I have my moles of KCl determined, I can plug it into the molarity equation and determine the liters and turn it into milliliters like the problem asks.

0.100 M KCl=(0.02612 mol KCl/x)

solve for x, x= 0.261 liters.

(0.261 liters)(1000 milliliters)= 261 mL of 0.100 M KCl


Thanks for looking over this, I hope I got it right!

[chenbeier]

Yes everything looks fine.

[potatoeater567]

Yes everything looks fine.

Thanks! I appreciate you going through my work and confirming it.