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

Answer:

0.001970 moles of sodium oxalate are present in the flask.

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:

$n=\frac{0.2640 g}{134 g/mol}=0.001970 mol$

0.001970 moles of sodium oxalate are present in the flask.