A 0.8935 g sample of an iron (ii) compund in dissolved in water and its titrated with KMnO4. if this titration requires 17.96 ml of 0.02515M KMnO4 to reach the pink end point (complete reaction) what is percent by mass of Fe in the sample? write down the answerr in 1 decimal place , followed by symbol percent %

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A 0.8935 g sample of an iron (ii) compund in dissolved in water and its titrated with KMnO4. if this titration requires 17.96 ml of 0.02515M KMnO4 to reach the pink end point (complete reaction) what is percent by mass of Fe in the sample? write down the answerr in 1 decimal place , followed by symbol percent %

Chemistry
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\(\Large Percent ~by~ mass=\frac{g~Solute}{g~Solution}\) Solute is Fe and solution is Fe + KMnO4 Here, you are given the titration in which it requires 17.96 ml of 0.02515M KMnO4 By this given info, convert ml to liters then (by molarity) convert liters to mols \(Molarity=\Large \frac{Mols}{Volume}=\frac{mols}{Liters}\)
The first step is to convert ml to liters The second step is to convert from liters to mols the third step is to convert from mols to grams, and there you have your mass for KMnO4

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We are solving for KMnO4's mass because we don't have the mass for it but we do for Fe (iron II). We can't solve the percent mass if we don't have the mass for KMnO4. \(Mass~Percent=\Large\frac{\color{blue}{g~Fe(Iron)}}{\color{blue}{g~Fe(Iron)}+\color{red}{KMnO4}}\) grams of KMnO4 is missing
@Zale101 i noticed that we can derive gram of KMnO4 from the Molarity and Volume that was given above? we can use that right to find the mass of KMnO4?
Correct.
Great explanation @Zale101 ! :)
You can derive grams of the given ml and molarity of KMnO4 because molarity will lead you to mols and from mols you can convert from mols to grams by using the molar mass.
Thanks @vera_ewing !
yesss..thank you so much Zale!!! im doing my assignment with this quest! @Zale101
No problem! :)

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