Question

How to find Moles of KMnO₄to neutralize C₂O₄^–   here is the equation :

5C₂O₄^2-+2MnO₄+16H Þ10CO₂+ 2Mn²⁺+H₂O

this is my data:

Green salt lab data

WEEK 1- 10.1159g ferrous ammonium sulfate

WEEK 2- weight of weighing bottle- 4.7942g

Weight of weighing bottle with crystals- 9.1301g

Weight of sodium oxalate- .1211g- trial 1 .1213g- trial 2

Trial 1 – initial buret reading- 1.99mL

Final buret reading- 19.00mL

Trial 2- initial buret reading- .95mL

Final buret reading- 18.25mL

Blank trial- initial buret reading- .99mL

Final buret reading- 1.00mL

Answer 1

The balanced equation is

5C₂O₄(aq) + 2MnO₄(aq) + 16H(aq)   10CO₂(g) + 2Mn(aq) + 8H₂0(l)

Here,

5 moles of Na₂C₂O₄ are neutralized by 2 moles of KMnO4.

1 moles of Na₂C₂O₄ are neutralized by 2/5 moles of KMnO4.

Now, Molar mass of Na₂C₂O₄ = 134 g/mol

So, 134 g of Na₂C₂O₄ = 1 mol

1 g of Na₂C₂O₄ = 1/134 mol

0.1211 g of Na₂C₂O₄ = (0.1211/134) mol = 0.0009 mol

and

0.1213 g of Na₂C₂O₄ = (0.1213/134) mol = 0.0009 mol

Since, 1 moles of Na₂C₂O₄ are neutralized by 2/5 moles of KMnO4.

0.0009 moles of Na₂C₂O₄ are neutralized by {(0.0009) (2/5)} moles of KMnO4.

0.0009 moles of Na₂C₂O₄ are neutralized by 0.00036 moles of KMnO4.