10) D

To solve this question, we need to find out the moles of lead nitrate solution and add this value to the moles of potassium iodide solution.

Formula:
Moles = Volume x Concentration

Just by looking at the values given in the table, we can see that D would give the largest value.

However, we will do the working to show you how to obtain D.

A = n[lead(II) nitrate sol] + n{2 x potassium iodide sol]
       (0.005×2) + (0.01 x 2 x 2) = 0.03 mol

B = (0.0025×5) + (0.0025 x 2.5 x 2) = 0.0375 mol

C= (0.0075×3) + (0.005 x 5 x 2) = 0.0725 mol

D= (0.005×4) + (0.0075 x 5 x 2) = 0.095 mol