Determine the weight of Na2C2O4 which is required to react completely with 20.0ml of 0.02M KMnO4.

Write and balance the equation.

2MnO4 + 5C2O4^2- ==> 10CO2 + 2Mn^2+ and you can finish but the coefficients are are what you need.
How much KMnO4 do you have? That's mols = M x L = 0.02L x 0.02M = ?
Use the coefficients above to convert mols KMnO4 to mols Na2C2O4.
Now convert mols Na2C2O4 to grams.
g = mols x molar mass.

To determine the weight of Na2C2O4 required to react completely with 20.0 mL of 0.02 M KMnO4, we'll need to use the concept of stoichiometry.

Step 1: Write the balanced chemical equation for the reaction between Na2C2O4 and KMnO4:

2 KMnO4 + 5 Na2C2O4 + 8 H2SO4 → K2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O + 5 Na2SO4

Step 2: Calculate the number of moles of KMnO4:

Given: Volume of KMnO4 solution = 20.0 mL = 0.020 L
Molarity of KMnO4 solution = 0.02 M

Number of moles of KMnO4 = volume x molarity
= 0.020 L x 0.02 mol/L
= 0.0004 mol

Step 3: Use the balanced chemical equation to determine the mole ratio between KMnO4 and Na2C2O4:

From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of Na2C2O4.

Step 4: Calculate the number of moles of Na2C2O4 required:

Using the mole ratio, we can set up the following proportion:

(0.0004 mol KMnO4) / (2 mol KMnO4) = (x mol Na2C2O4) / (5 mol Na2C2O4)

Cross-multiply and solve for x:

x = (0.0004 mol KMnO4) x (5 mol Na2C2O4) / (2 mol KMnO4)
= 0.001 mol Na2C2O4

Step 5: Calculate the weight of Na2C2O4:

Given: Molar mass of Na2C2O4 = 134.00 g/mol

Weight of Na2C2O4 = number of moles x molar mass
= 0.001 mol x 134.00 g/mol
= 0.134 g

Therefore, the weight of Na2C2O4 required to react completely with 20.0 mL of 0.02 M KMnO4 is 0.134 grams.