Hello, this is an Ap chemistry student and I really need help, could anyone plz help me?

1. Write a balanced equation

2AgNO3(aq) + CaCl2(aq) --------> 2AgCl(s) + Ca(NO3)2(aq)

2. Calculate the number of moles of each reagent you are starting with

moles = molarity x L

moles AgNO3 = 0.20M x 0.1000 L
moles AgNO3 =  0.020 moles

moles CaCl2 = 0.15 M x 0.1000 L
moles CaCl2 = 0.015 moles

3. Work out which is the limiting reagent. The limiting reagent is the one that will be all used up in the reaction. There is not enough limiting reagent to use all the other reagent so the other reagent is called the excess reagent.

Your balanced equation tells you that
1 mole CaCl2 needs 2 moles AgNo3 to fully react
Therefore 0.015 moles CaCl2 needs (2 x 0.015) moles AgNO3
= 0.030 moles AgNO3 needed to react all the CaCl2

But you only have 0.020 moles AgNO3, which is not enough, therefore AgNO3 is the limiting reagent.

4. Calculate the moles of product formed. The maxiimum amount of product you can get is if all the limiting reagent reacts fully. So use the moles of limiting reagent to determine the moles of AgCl you will make.

Balanced equation tells you that
2 moles AgNO3 react to give 2 moles AgCl
Therefore 0.020 moles AgNO3 will react to form 0.020 moles AgCl

mass = molar mass x moles
mass AgCl = 143.32 g/mol x 0.020 moles
mass AgCl = 2.8664 g
= 2.9 g (2 sig figs)

5. Work out how much of the excess reagent was actually used

Balanced equation shows that

2 moles AgNO3 needs 1 mole CaCl2 to react
So 1 mole AgNO3 needs 1/2 mole CaCl2
Thus 0.020 moles AgNO3 needs (1/2 x 0.020) moles CaCl2
= 0.010 moles CaCl will be used up

Since you start with 0.015 moles CaCl2 then
left over CaCl2 = moles given - moles used = 0.015 mol - 0.010 mol
= 0.005 moles CaCl2 unused

Now, each 1 CaCl2 gives 1 Ca2+ ion and 2 Cl- ions
CaCl2 ----------> Ca^2+  + 2Cl-

So moles Ca2+ = 0.005 moles
moles Cl- = 0.010 moles

7. Calculate concentration form moles and volume

Total volume of solution = 100.0 ml + 100.0 ml = 200.0 ml
= 0.2000 L

Molarity = moles / L

Molarity Cl- = 0.010 mole / 0.2000 L
= 0.050 M

Molarity Ca2+ = 0.0050 mol / 0.2000 L
= 0.025 M