Question

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?

Answer 1

678 mL.

Step-by-step 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 =
`(mol)/(M)`

M = 0.36 M

mol = 0.244

Putting values

volume = 0.678 L = 678 mL.