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

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)

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.

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

To calculate the theoretical yield and percent yield, you need to know the molar masses of the compounds involved.

The molar mass of AgNO3 is 170g/mol. Therefore, using the given mass of 2AgNO3 (1.5g), you can calculate the number of moles of AgNO3:

moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 1.5g / 170g/mol
≈ 0.009 moles

Since the stoichiometry of the balanced equation is 2:2, this means that for every 2 moles of AgNO3, there will be 2 moles of AgCl produced:

moles of AgCl = moles of AgNO3
= 0.009 moles

Now, you can calculate the mass of AgCl using its molar mass of 143g/mol:

mass of AgCl = moles of AgCl * molar mass of AgCl
= 0.009 moles * 143g/mol
≈ 1.287g

So, the theoretical yield of AgCl is approximately 1.287g.

To calculate the percent yield, you need to know the actual yield. However, in this case, the actual yield is not given. Therefore, you cannot determine the percent yield.

To determine the limiting and excess reactant, compare the number of moles of each reactant to their stoichiometric ratio in the balanced equation. The reactant that will produce less product is the limiting reactant, while the other reactant is in excess.

For AgNO3:
moles of AgNO3 = 0.009 moles

For CaCl2:
moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
= 1.1g / 110g/mol
= 0.01 moles

Since the stoichiometric ratio between AgNO3 and CaCl2 is 2:1, it is clear that AgNO3 is the limiting reactant because it produces only 0.009 moles of AgCl, whereas CaCl2 could potentially produce 0.01 moles of AgCl.

To calculate how much excess remains, you can subtract the number of moles of AgCl produced by the limiting reactant from the number of moles of AgCl that could potentially be produced by the excess reactant (in this case, CaCl2).

moles of excess = moles of CaCl2 - moles of AgCl (produced by the limiting reactant)
= 0.01 moles - 0.009 moles
= 0.001 moles

Now, you can calculate the mass of excess remaining using the molar mass of CaCl2:

mass of excess = moles of excess * molar mass of CaCl2
= 0.001 moles * 110g/mol
= 0.11g

So, approximately 0.11g of excess CaCl2 remains.