What volume of this 0.200 mol/L lead nitrate solution is just sufficient to react with all the iodide in 25mL of 0.247 mol/L potassium iodide solution?
First, write out the equation. (going to be ignoring states as it would be annoying)
Pb(NO3)2 + 2KI -> 2KNO3 + PbI2
Number of moles of potassium iodide solution you have:
0.247 x 0.025 = 6.175 x 10^-3 moles
Mole ratio of lead nitrate to potassium iodide is 1:1
So you have the same amount of moles, now calculate volume.
Volume = (6.175 x 10^-3)/0.200
= 0.031L (2sf)