Question

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.
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 = ?

Related Questions

Stoichiometry 0.083grams of lead nitrate, Pb(NO3)2, in solutions mixed with 0.300g of sodium iodi... Determine themass of the copper solid produced when excess magnessuim metal reacts with 250 ml of 0.... A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide... We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding pot...