Write a correctly balanced equation for the reaction:
MgCl2(aq) + 2AgNO3(aq) ==> Mg(NO3)2(aq) + 2AgCl(s)
From the net ionic equation:
2Ag+(aq) + 2Cl-(aq) ==> 2AgCl(s)
From the balanced equation, it can be seen that it takes ONE mole MgCl2 to precipitate TWO moles of AgNO3 in the form of AgCl.
So, first calculate how many moles of Ag you have in 25.0 ml of 0.25 M AgNO3:
0.025 L x 0.25 moles AgNO3/L x 1 mole Ag/mole AgCl = 0.00625 moles Ag present
Next, calculate moles of MgCl2 needed to precipitate 0.00625 moles Ag:
0.00625 moles Ag x 1 mole MgCl2/2 moles Ag = 0.003125 moles MgCl2
Finally, find the volume of 0.15 M MgCl2 needed to supply this amount:
(x L)(0.15 moles/L) = 0.03125 moles and x = 0.02083 liters = 20.8 ml = 21 ml (2 significant figures).