Question

3.0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium permanganate to titrate it to turn pink end point. The equation for this reaction is: 5Na,C,O4+ 2KMnO,+ 8H,SO 2MnSOg+ K,SO,+ 5 Na,SO,+ 10CO2+ 8H2O a) How many moles of sodium oxalate are present in the flask?

Answer 1

0.001970 moles of sodium oxalate are present in the flask.

Step-by-step explanation:

Mass of sodium oxalate = m = 0.2640 g

Molar mass of sodium oxalate =
`Na_2C_2O_4` = 134 g/mol

Moles of compound:n

`n=\frac{\text{Mass of compound}}{\text{Molar mass of compound}}`

Number of moles of sodium oxalate:

`n=(0.2640 g)/(134 g/mol)=0.001970 mol`

0.001970 moles of sodium oxalate are present in the flask.