What is the maximum volume of a 0.788 m cacl2 solution that can be prepared using 85.3 g cacl2?

The answer is 0.975 L

Volume = mol/Molarity

We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:

Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol

So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles

Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L

On the reactant side you get 
2 Cl from BaCl₂

And on the product side you get 
2 Cl from 2 NaCl

∴ the answer would be 2 on the reactant and 2 on the product