It si about chemistry practical

Question

1. What volumen of 0.0200 moldm3 kmno4 solution is required to oxidixed 40.0cm3 of 0.100moldm3 feso4 un sulfuric acid solution.

Answer 1

I'll rewite the question to try and clarify the language and units:

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1. What volume of 0.0200 mole/liter KMnO4 solution is required to oxidize 40.0 ml of 0.100mole/liter FeSO4 and sulfuric acid solution?

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I interpreted "moldm3" as moles/dm^3.

1 dm^3 is 1 liter

1 cm^3 is 1 ml.

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Assuming my rewrite is correct:

We need a balanced equation for this reaction. I find the following.

2KMnO4​+8H2​SO4​+10FeSO4​→K2​SO4​+2MnSO4​+5Fe2​(SO4​)3​+8H2​O

We are asked how much of the KMnO4 reactant is needed to oxidize the FeSO4 reactant. We may assume the H2SO4 is not a limiting reagent in this reaction.

The balanced equation tells us that 2 moles of KMnO4 are quired for every 10 moles of FeSO4, which is a molar ratio of 2/10 (moles KMnO4)/(moles K2SO4).

Lets calculate the moles of FeSO4 are present in 40 ml of 0.100M FeSO4.

Remember that Molarity is defined as moles/liter. 0.100 M means there are 0.100 moles of solute per liter.

Volume (in L) x Molarity(moles/liter) = moles

40 ml is 0.04 liters

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  (0.040L)*(0.100 moles/liter) = 0.004 moles FeSO4

To find the moles of KMnO4 that are required, multiply the moles FeSO4 times the molar ratio calculated earlier:

 

  (0.004 moles FeSO4)*[2/10 (moles KMnO4)/(moles K2SO4).] = 0.00040 moles of KMnO4

We need a volume of the 0.02 M KMnO4  reagent that provides 0.00040 moles of KMnO4.

Volume (in L) x Molarity(moles/liter) = moles

Volume (in L) x (0.02 moles/liter) = 0.00040 moles

Volume of KMnO4 = 0.020 liters, or 20 ml

20 ml of 0.02 M KMnO4 is needed to oxidize 40.0 ml of a 0.100 M FeSO4 solution.