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8_murik_8 [283]
3 years ago
12

Potassium permanganate(KMnO4) reacts with oxalic acid (H2C2O4) in aqueous sulfuric acid according to: 2KMnO4 5H2C2O4 3H2SO4->

2MnSO4 10 CO2 * H2O K2SO4 How many milliliters of a .250 M KMnO4 solution are needed to react completely with 3.225 G of oxalic acid
Chemistry
1 answer:
Andru [333]3 years ago
5 0

Answer:

57.3mL of 0.250M KMnO4 are required

Explanation:

3.225g of oxalic acid are:

3.225g×(1mol / 90.03g) = <em>0.03582mol of oxalic acid</em>

Based on the reaction, 5 moles of oxalic acid react with 2 moles of KMnO4, thus, for a complete reaction of oxalic acid you need:

0.03582mol of oxalic acid × (2mol KMnO4 / 5mol Oxalic Acid) = <em>0.01433mol of KMnO4</em>

<em />

If concentration of KMnO4 is 0.250M, liters in 0.01433mol are:

0.01433mol × (1L /0.250mol) = 0.0573L ≡<em> 57.3mL of 0.250M KMnO4 are required</em>

<em></em>

I hope it helps!

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