A 20.00 mL sample of MnO4 is required to titrate .2378 g Na2C2O4 in an acidic solution. How many mL of this same MnO4 -1 are required to titrate a 25.00mL sample of 0.1010 M Fe+2 in acidic solution

We need 2.85 mL of MnO4-

Explanation:

Step 1: Data given

Volume of MnO4 = 20.00 mL = 0.020 L

Mass of Na2C2O4 = 0.2378 grams

Molar mass Na2C2O4 = 134 g/mol

Volume of a Fe^2+ solution = 25.00 mL

Molarity of Fe^2+ = 0.1010 M

Step 2: The balanced equation

2MnO4 + Na2C2O4 → C2O4 + 2NaMnO4

Step 3: Calculate moles Na2C2O4

Moles Na2C2O4 = mass Na2C2O4 / molar mass Na2C2O4

Moles Na2C2O4 = 0.2378 grams / 134 g/mol

Moles Na2C2O4 = 0.00177 moles

Step 4: Calculate moles MnO4

For 1 mol Na2C2O4 we need 2 moles MnO4

For 0.00177 moles Na2C2O4 we need 2*0.00177 = 0.00354 moles MnO4

Molarity MnO4 = 0.00354 moles / 0.02 L

Molarity MnO4 = 0.177 M

Step 5: The balanced equation

5 Fe^2+ + MnO4- +8H+ → 5Fe^3+ + Mn^2+ + 4H2O

Calculate volume of MnO4

b * Ca*Va = a * Cb*Vb

⇒with b = the coefficient of MnO4 = 1

⇒with Ca = the concentration of Fe^2+ = 0.1010 M

⇒with Va = the volume of Fe^2+ = 25.00 mL = 0.025 L

⇒with a = the coefficient of Fe^2+ = 5

⇒with Cb = the concentration of MnO4- = 0.177 M

⇒with Vb = the volume of MnO4 = TO BE DETEMRMINED

0.1010 * 0.025 = 5 * 0.177 Vb

Vb = 0.00285 L = 2.85 mL

We need 2.85 mL of MnO4-

answer; /// i do believe that the correct answer is (c)nuclear bond