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Volume of 0.1 m ferrous oxalate solution required to react completely with 60 ml of 0.1 n acidified kmno4 solution is :

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Answer:

100 ml of the 0.1 M ferrous oxalate solution.

Step-by-step explanation:

Ferris Oxalate is FeC2O4, an inorganic compound.

The reaction with KMnO4 occurs in an acidified solution. We'll assume sulfuric acid (H2SO4) for this reaction.

FeC2O4 + KMnO4 + H2SO4 → Fe2(SO4)3 + CO2 + MnSO4 + K2SO4 + H2O

This needs to be balanced. Start with the most complex molecule and assign a "1" as it's coefficient. That means we need 2 Fe atoms, so make the coefficient of FeC2O4 a "2."

The rest is not as easy. Each atom must be accounted for, which will required several iterations until the balanced equation is found.

Without explaining that [painful] process, the following equation appears balanced.

10 FeC2O4 + 6 KMnO4 + 24 H2SO4 → 5 Fe2(SO4)3 + 20 CO2 + 6 MnSO4 + 3 K2SO4 + 24 H2O

The need to balanced the chemical equation is to establish how many moles are needed for both the KMnO4 and FeC2O4 reactants for a complete reaction (both reactants are entirely consumed).

The balanced equation says that 10 moles of FeC2O4 will react completely with 6 moles of KMnO4. That is a 10/6 or 5/3 molar ratio.

Once the number of moles of KMnO4 is determined, it should be multiplied by (5/3) to find the number of moles of FeC2O4 that would be consumed.

The number of moles of KMnO4 in 60 ml of 0.1 N acidified KMnO4 solution can be calculated:

For KMnO4, 0.1 N is the same as 0.1 M.

The definition of M is moles/liter. ).1 M means the solution has 0.1 moles KMnO4 per liter of solution.

The number of moles of KMnO4 is therefore:

(60 ml)*(0.001 L/ml)*(0.1 moles/L) = 0.006 moles KMnO4

From above, the balanced equation told us that 10 moles of FeC2O4 will react completely with 6 moles of KMnO4, a 5/3 molar ratio (FeC2O4/KMnO4).

The moles of FeC2O4 required to react completely with 0.006 moles KMnO4 is therefore:

(0.006 moles KMnO4)*(5/3 (FeC2O4/KMnO4)) = 0.01 moles FeC2O4

Now determine the number of ml of a 0.1 M ferrous oxalate solution that would contain 0.01 moles of FeC2O4. Let X be the volume (L) of 0.1 M ferrous oxalate solution required:

X*(0.1M) = 0.01 moles FeC2O4

X*(0.1 moles/L) = 0.01 moles FeC2O4 [M is moles/L]

X = (0.01 moles FeC2O4)/(0.1 moles/L)

X = 0.1 L FeC2O4

X = 100 ml (0.1M FeC2O4 solution)

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