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### Topic: oxidation and reduction calculations  (Read 3408 times)

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#### pat065

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##### oxidation and reduction calculations
« on: February 01, 2011, 02:33:45 PM »
what volume of 0.0200 mol dm-3 of tin(II) nitrate will be oxidised by 25cm3 of 0.0200mol dm-3 potassium manganate(VII) in acid solution?

Atm i have got the equations:

KmNO4+8H+5e- = Mn2+ + 4H2O and Sn2+ = Sn4+ +2e-

Therefore ratio is 5:2

which means 2.5 times

but then how do i do this, i have a mock in a couple of days and so i am trying to understand these sorts of questions!

If u could help it would be much appreciated  thanks

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##### Re: oxidation and reduction calculations
« Reply #1 on: February 01, 2011, 02:37:33 PM »
well, you are right about the 2.5 times.

but let's start at the start: how many moles of KMNO4 do you have in your solution?

then you can use the 2.5 to determine how much tin you then have oxidised.

then you convert the moles of tin to volume of your 0.0200M solution

can you work this out?

#### pat065

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##### Re: oxidation and reduction calculations
« Reply #2 on: February 01, 2011, 03:01:49 PM »
mole of KMNO4. does this mean using so that means moles= conc x volume

which= 0.0200 mol dm-3 x 25cm3 = 0.5moles

so then do u times 0.5/1000 x 2.5 =0.00125

then input into v=n/m

therefore 0.00125/0.0200= 1/16

then x by 1000
so u need 62.5cm3 of the tin solution

#### vmelkon

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##### Re: oxidation and reduction calculations
« Reply #3 on: February 05, 2011, 10:07:17 PM »
dm^3 is liters. Don't type dm-3
cm^3 is mL.

You basically wrote this
0.0200 mol/L x 25 mL = 0.5moles

You would get 0.5 mol if you had 25 L!

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##### Re: oxidation and reduction calculations
« Reply #4 on: February 07, 2011, 01:01:15 PM »