This equation decribes oxidation of acidified solution of sodium oxalate(Na2C2O4) by potassium permanganate(KMnO4):
5 C2O4^2- + 2Mn04^- + 16H2SO4 to produce 10Co2 + 2Mn^2- + 16HSO4^- + 8H2O.
what volume of 0.127M sodium oxalate in sulphuric acid will react with 3.57g of potassium permanganate?
I got 177ml as my answer. Not sure if it is right. Pls help A.S.A.P .
Thanks
I didn't get 177 mL.
mols KMnO4 = grams/molar mass
mols Na2C2O4 = mols KMnO4 x (5 mols ox/2 mol MnO4) = ?
Them M Na2C2O4 = mols/L
Solve for L and convert to mL.
To find the volume of 0.127 M sodium oxalate that will react with 3.57 g of potassium permanganate, we need to find the number of moles of potassium permanganate first.
The molar mass of KMnO4 = 39.1 g/mol (K) + 54.9 g/mol (Mn) + 16.0 g/mol (O) x 4 = 158.04 g/mol.
Number of moles of KMnO4 = mass / molar mass = 3.57 g / 158.04 g/mol ≈ 0.0226 mol.
From the balanced equation, we can see that 2 moles of KMnO4 are required to react with 5 moles of sodium oxalate (Na2C2O4).
Therefore, the number of moles of sodium oxalate needed = (0.0226 mol x 5) / 2 = 0.0565 mol.
Now, let's calculate the volume of 0.127 M sodium oxalate solution needed:
Molarity (M) = moles / volume (L).
Rearranging the equation, volume (L) = moles / Molarity.
Volume of sodium oxalate solution = 0.0565 mol / 0.127 mol/L = 0.445 L = 445 mL.
Therefore, the volume of a 0.127 M sodium oxalate solution required to react with 3.57 g of potassium permanganate is 445 mL, not 177 mL.
Please check your calculations again to see if there was an error.
To determine the volume of the sodium oxalate solution required, we can use the concept of stoichiometry and the given amount (mass) of potassium permanganate.
First, let's calculate the number of moles of potassium permanganate (KMnO4) using its molar mass:
Molar mass of KMnO4 = 39.1 (K) + 54.9 (Mn) + 4 x 16 (O) = 158.04 g/mol
Number of moles of KMnO4 = mass / molar mass = 3.57 g / 158.04 g/mol ≈ 0.0226 mol
Now, let's use the stoichiometry of the equation to find the number of moles of sodium oxalate (Na2C2O4) required. From the equation, we can see that the ratio of sodium oxalate to potassium permanganate is 5:2:
Number of moles of Na2C2O4 = (0.0226 mol KMnO4) x (5 mol Na2C2O4 / 2 mol KMnO4) = 0.0565 mol Na2C2O4
Finally, we can calculate the volume of the sodium oxalate solution using its molarity and the number of moles:
Molarity of sodium oxalate solution (M) = 0.127 M
Volume of sodium oxalate solution (V) = number of moles / molarity = 0.0565 mol / 0.127 mol/L = 0.445 L = 445 mL
Therefore, the volume of the 0.127 M sodium oxalate solution required to react with 3.57 g of potassium permanganate is approximately 445 mL, not 177 mL.