Final answer:
The correct volume of 0.100 M KI needed to react with 40.0 mL of 0.0400 M Hg2(NO3)2 is 64.0 mL, which is not properly matched by any of the given options. The closest choice is 60.0 mL.
Step-by-step explanation:
To determine how many milliliters of 0.100 M KI are needed to react with 40.0 mL of 0.0400 M Hg2(NO3)2, we need to start by writing the balanced chemical equation for the reaction, which is somewhat incorrectly provided in the question. It should be Hg2(NO3)2 + 4KI → 2HgI2 + 2KNO3. According to this correct equation, one mole of Hg2(NO3)2 reacts with four moles of KI. We need to calculate the number of moles of Hg2(NO3)2 in 40.0 mL. Since the concentration is 0.0400 M, we have 0.0400 mol/L × 0.0400 L = 0.00160 mol of Hg2(NO3)2. Now, for each mole of Hg2(NO3)2, we need four times as many moles of KI, so we need 0.00160 mol × 4 = 0.00640 mol of KI. To find the volume of 0.100 M KI needed, we use the concentration of KI (CKI = 0.100 mol/L) and the number of moles required (nKI = 0.00640 mol). Volume (V) can be calculated by rearranging the molarity equation n = C × V, giving us V = n / C. Therefore, V = 0.00640 mol / 0.100 mol/L = 0.0640 L, which is 64.0 mL of KI. However, none of the provided options match this result, so it appears there is an error in the options. The closest would be Option 4: 60.0 mL.