Respuesta :
Answer:
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-
