1.Calculate the Theoretical and % Yields of the following equation:

Question

2.What is the limiting and excess reactant?
3.How much excess remains?

2AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2
mass of 2AgNO3= 1.5g
mass of CaCl2= 1.1g

(What I have so far):
AgNO3:
1.5g(AgNO3) 2mol/170g = 0.018mol(AgNO3)

0.018mol(AgNO3) 2mol(AgCl)/2mol(AgNO3) = 0.018mol(AgCl)

0.018mol(AgCl) 143g/2mol = [1.287g(AgCl)] <-Theoretical Yield
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CaCl2:
1.1g(CaCl2) 110g/1mol = 121mol(CaCl2)

121mol(CaCl2) 2mol(AgCl)/1mol(CaCl2) = 242mol(AgCl)

242mol(AgCl) 143g/2mol = 17303g(AgCl)

Answer 1

You're on the right track but you have your numbers a little mixed up. moles = grams/molar mass.
moles AgNO3 = 1.5/170 = 0.00882
0.00882 moles AgNO3 x (2 mol AgCl/2 mol AgNO3) = 0.00882 mol AgCl produced (trial).
moles CaCl2 = 1.1/111 = 0.00991
You used the wrong molar mass and multiplied instead of dividing
0.00991 mol CaCl2 x (2 mol AgCl/1 mol CaCl2) = 0.00991*2 = 0.0198 mol AgCl produced.(trial)
In limiting reagent problems the SMALLER value is ALWAYS the correct one to choose(both answers obviously can't be right so you choose the smaller one).
So 0.00882 mol (it isn't necessary to change to grams to know this) is the one to choose and that will be the limiting reagent. That is the AgNO3 so CaCl2 must be the reagent in excess.

To find how much excess CaCl2 is present one must determine how much CaCl2 was used. That's just another stoichiometry problem. Since all of the AgNO3 will be used, we use it to start.
0.00882 mol AgNO3 x (1 mol CaCl2/2 mol CaCl2) = 0.00441 mol CaCl2 used. That is how many grams.? 0.00441 x 111 = 0.04895g used. Therefore, 1.10 - 0.05 = 0.6 g CaCl2 remain unreacted.
Check my work.

Answer 2

Consider the following reaction. If 5 grams of Calcium Chloride react with 1.23 moles of Silver Nitrate, calculate: a) Mass of silver Chloride formed.
b) Excess Reactant