Question

Answer the following questions based on this reaction:
Pb(NO3)2 (aq) + 2 NaCl (aq) --> PbCl2 (s) + 2 NaNO3 (aq)
a) If 225 mL of 12.95 M Pb(NO3)2(aq) are reacted with a solution made with 5.05 g of NaCl (aq), how many grams of lead (II) chloride will be precipitated

b) If the actual yield of lead (II) chloride is 1.06 g, what is the percent yield?

PLEASE HELP :( I HAVE MY FINAL IN A COUPLE WEEKS!

Answer 1

This is another LR problem. Look at the last problem to see the process used. For mols of solutions, mols = M x L = ? while that of solids is mols = grams/molar mass. When you find the mass PbCl2 that is the theoretical yield.

Then yield is
% yield = (actual yield/theoretical yield)*100 = ?
Post your work if you get stuck.