Determine the mass of the precipitate, lead iodide that is formed when 150 ml of 0.0500M pb(no3)2 is reacted with excess potassium iodide. if 3.3 of the precipitate is experimentally recovered, determine the percent yeild.

Answer 1

Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
mols Pb(NO3)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols PbI2 ppt.
Now convert mols PbI2 to grams PbI2. That's grams = mols x molar mass = ? That is the theoretical yield (TY). The actual yield (AY) is 3.3g.

% yield = (AY/TY)*100 = ?