Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Answer:
The correct answer is option B
Explanation:

Given values,
Molarity of 
Volume of solution, 
Molecular weight of 
Substituting this values in Molarity formula, we get

D. Kinda simple, don’t think much explaining is needed.
Answer:
u will equate and make V2 the subject because it's the one u are looking for ....