Stoichiometry and Percent Yield

Question

A student reacted 0.500 g of Pb(NO₃)₂ with 0.750 g of KI according to the reaction,

Pb(NO₃)₂ (aq) + 2 KI (aq) --> PbI₂ (s) + 2 KNO₃ (aq)

If the student obtained 0.538 grams of PbI₂ product after collecting it by filtration and drying, what was the percent yield of PbI₂ obtained? _____ %

If the student calculated a 90.7% yield after performing the experiment above, what was the amount of PbI₂ product formed in his experiment? ____  g PbI₂

Answer 1

Pb(NO₃)₂ (aq) + 2 KI (aq) --> PbI₂ (s) + 2 KNO₃ (aq) ... balanced equation

To find % yield, one must first find the theoretical yield.

Theoretical yield:

Find the limiting reactant.

moles Pb(NO3)2 = 0.500 g Pb(NO3)2 x 1 mol / 331 g = 0.00151 mols

moles KI = 0. 750 g KI x 1 mol / 166 g = 0.00452 mols

Limiting reactant is Pb(NO3)2 based on the mol ratio of 1:2 in the balanced equation.

Theoretical yield PbI₂ = 0.00151 mol Pb(NO₃)₂ x 1 mol PbI₂/mol Pb(NO₃)₂ x 461 g PbI₂/mol = 0.696g PbI₂

Percent yield:

% yield = actual yield / theoretical yield (x100%)

% yield = 0.538 g / 0.696 g (x100%) = 77.3% yield

x g / 0.696 g = 0.907

x = 0.631 g PbI₂

Answer 2

Please use the following link to the document to find the worked out solution. (WyzAnt is giving me an error when I try to submit my answer on here).

https://docs.google.com/document/d/1FJunYlSYqIBGAdMW0M5ur5xOa8QuavfaIs-cAXjrhF0/edit?usp=sharing