What mass of precipitate is formed when 0.750 L of 2.50 mol/L CaCl2 reacts with 650 mL of 1.75 mol/L AgNO3

Answer 1

You have a limiting reagent problem here. You know that because amounts are give for BOTH of the reactants.

2AgNO3 + CaCl2 ==> 2AgCl(s) + Ca(NO3)2

mols AgNO3 = M x L = ?
mols CaCl2 = M x L = ?

Using the coefficients in the balanced equation, convert mols AgNO3 to mols AgCl.
Do the same and convert mols CaCl2 to mols AgCl.
It is likely that the two values will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller one and the reagent responsible for that is the LR.

Using the smaller value for mols AgCl, convert to grams. g AgCl = mols AgCl x molar mass AgCl.

Answer 2

How do i convert mols cacl2 to agcl

Answer 3

You use the coefficients in the balanced equation to convert anything you wish. For CaCl2 to mols AgCl it is
?mols CaCl2 x (2 mols AgCl/1 mol CaCl2) = ?mols CaCl2 x 2/1 = ?mols AgCl.

Answer 4

I don't understand...

Answer 5

How did you convert mols AgNO3 to mols AgCl?