Find the theoretical and actual yield of 25.0g Pb(NO3)2+15.0g2NaI=PbI2+2Na(NO2)

Answer 1

You made a typo in the equation; it should be
Pb(NO3)2 + 2NaI ==> PbI2 + 2NaNO3 (should be NO3 and not NO2 with NaNO3. Also, note that you don't put quantities in the equation line.

This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
mols Pb(NO3)2 = grams/molar mass = ?
mols NaI = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols PbI2.
Do the same for converting mols NaI to mols PbI2
You will likely get different values for mls PbI2 which means one of the values is wrong. The correct value in LR problems is ALWAYS the smaller one and the reagent providing that smaller number is the LR.
Then convert mols PbI2 to grams by grams = mols x molar mass. That is the theoretical yield for PbI2. If you want theoretical yield for NaNO3 follow the procedure above to calculate that.
You can't calculate the actual yield unless you know the percent yield and that isn't given in the problem.