Pb(NO3)2 + -
(aq) + 2 Nal (aq) --> Pbl2 (s) + 2 NaNO3 (aq)
Starting with with 50.0 grams...
Pb(NO3)2 + -
(aq) + 2 Nal (aq) --> Pbl2 (s) + 2 NaNO3 (aq)
Starting with with 50.0 grams of Pb(NO3)2 and 30.0 grams of Nal:
A. What is the limiting reagent?
B. How many grams of the excess reactant remains?
C. How many grams of each product is formed?
D. If 12 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Answers: 2
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