Final answer:
To calculate the necessary volume of 0.0100 M MnO4− to react with 0.637 g of FeCl2, convert the mass of FeCl2 to moles, use the stoichiometry of the reaction to find the moles of MnO4− needed, and then divide by the molarity of MnO4− to find the volume. The result is 100.4 mL.
Step-by-step explanation:
To calculate the volume of 0.0100 M MnO4− needed to react with 0.637g FeCl2, you first need to determine the moles of Fe2+ in the FeCl2. Using the molar mass of FeCl2 (which is 126.75 g/mol), you can calculate the moles of Fe2+ using the formula:
moles Fe2+ = mass of FeCl2 / molar mass of FeCl2
= 0.637g / 126.75 g/mol
= 0.00502 mol
We then need the balanced equation of the reaction between Fe2+ and MnO4− in acidic solution. For this example, let's assume the balanced equation is:
5 Fe2+ + MnO4− + 8 H+ → 5 Fe3+ + Mn2+ + 4 H2O
Next, from the balanced equation, there is a molar ratio of 5 moles of Fe2+ to 1 mole of MnO4−. Therefore, the moles of MnO4− needed are calculated by:
moles of MnO4− = moles of Fe2+ / 5 = 0.00502 mol / 5 = 0.001004 mol
Finally, to find the volume of 0.0100 M MnO4− needed, we use the equation:
V(MnO4−) = moles of MnO4− / Molarity of MnO4−
= 0.001004 mol / 0.0100 M
= 0.1004 L
= 100.4 mL
Therefore, 100.4 mL of 0.0100 M MnO4− is needed to react with 0.637g of FeCl2.