Consider that you have 330 ml of a 0.37 m aqueous solution of calcium chloride. what volume (in ml) of a 0.36 m aqueous solution of silver nitrate would you need in order to completely react with your aqueous calcium chloride solution?

678 mL.

Explanation:

The balanced reaction between calcium chloride and silver nitrate is shown in figure.

As shown in the equation , each mole of aqueous calcium chloride will rect with two moles of silver nitrate.

The moles of calcium chloride taken = molarity X volume (L)

                                                             = 0.37X0.33 = 0.122 mol

the moles of silver nitrate required = 2 X 0.122 = 0.244 mol

the volume of silver chloride required =

M = 0.36 M

mol = 0.244

Putting values

volume = 0.678 L = 678 mL.