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- samigupta8

how many milliliters of a 0.05 M kmno4 solution are required to oxidise 2 g of feso4 in a dilute acid solution

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- samigupta8

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- anonymous

well the answer is 52.57 mL of a .05 M KMnO4 solution, but let me get back to you with the explanation.

- samigupta8

okkkkk

- samigupta8

bt plssss.... make it fast for i hve to ask 1 more ques..

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- anonymous

ask the other question while im here, Molarity is always a tough subject to explain

- samigupta8

i'll ask it afterwrds firstly can u explain me how u got it

- samigupta8

plss....do tell me fast

- anonymous

first you find out how many moles there are in 2 grams of FeSO4. (2grams/molarmass)
->( 2g/151.908g/mol) = .0132 moles of FeSO4
-->now you divide .0132 moles of FeS04 by 5 because of the stoichiometric ratio between Fe2+ and Mn2+.
MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+
thats the balanced equation. There's a coefficient of 5 on Fe2+ therefore like i said before we divide our .0132moles of FeS04 by 5 so that it can give us an equal ratio to Mn2+
--> .0132moles/5 = .00264 moles of FeS04
The question asks us how many mL's of .05 Molar solution of KMnO4 are required to oxidize 2 grams of FeSO4. So far we converted the grams into moles, and in the end we got .00264 moles of FeSO4. Now we need the moles of KMnO4. Luckily they gave us its Molarity.
--> Molarity is (# of moles solute/1 L). Our molarity is .05 and anything divided by 1(Liter) is still the same number therefore we can conclude that the number of moles of KMn04 is .05.
We want mL's here so we are going to divide the number of FeSO4 moles by KMnO4 moles. --> .00264 moles / .05 moles =.0528
This .0528 has no units since the moles cancelled out in the last calculation therefore we
can multiply it by either 1L/or 1000mL. In our case our answer is looking for mL so we multiply
---> .0528 by 1000mL = 52.8 mL of .05 M KMnO4 is used to oxidize 2 grams of FeSO4.
If you have questions on the stoichiometric stuff i posted a link below:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method
hope that helped

- abb0t

Good explanation.

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