What volume in ml of 0.500 M silver nitrate is required to completely react with 25.0 ml of 1.5 M calcium chloride?

First, the balanced chemical equation of the reaction must be written. This gives:

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

This shows that for every mole of CaCl2, 2 moles of AgNO3 reacts (1:2 ratio). Given the volume and molarity of CaCl2, we can compute for the number of moles of CaCl2.

moles CaCl2 = 1.5M (25mL/1000) = 0.0375 moles

Now, we can also compute for the moles of AgNO3 using the balanced equation.

moles AgNO3 = 2 (0.0375) = 0.075 moles

Using the moles and molarity of AgNO3, we can compute for the volume needed.

volume (mL) / 1000 = 0.075/0.5 = 150 mL of silver nitrate solution is required.