# Chemistry A2 question struggled with

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#1
Can someone help me with this question my answer doesn't look right at all

Q: A student weighed out 6.30g of hydrated iron(II) sulphate, FeSO4.xH2O, and dissolved it in 50cm3 of sulfuric acid. This solution was poured into a 250cm3 volumetric flask and made up to the mark with distilled water. 25.0cm3 of this solution was titrated with 0.0200moldm-3 potassium manganate(VII), KMnO4, and 24.20cm3 of the MnO4- solution was used. The full ionic (redox) equation for the reaction is as follows:
MnO4- + 8H+ + 5Fe2+→ Mn2+ + 4H2O + 5Fe3+
Determine the value of x in the iron(II) salt.
0
3 years ago
#2
(Original post by Noogler)
Can someone help me with this question my answer doesn't look right at all

Probably best to also show your working.
0
3 years ago
#3
(Original post by Noogler)
Can someone help me with this question my answer doesn't look right at all

Q: A student weighed out 6.30g of hydrated iron(II) sulphate, FeSO4.xH2O, and dissolved it in 50cm3 of sulfuric acid. This solution was poured into a 250cm3 volumetric flask and made up to the mark with distilled water. 25.0cm3 of this solution was titrated with 0.0200moldm-3 potassium manganate(VII), KMnO4, and 24.20cm3 of the MnO4- solution was used. The full ionic (redox) equation for the reaction is as follows:
MnO4- + 8H+ + 5Fe2+→ Mn2+ + 4H2O + 5Fe3+
Determine the value of x in the iron(II) salt.
I'm assuming this is a 6-8 mark question. I'll try to explain each step of ccalculation.

1. n(MnO4-) = 0.02 x (24.2/1000) = 0.000484mol
We want to work out the number of moles of mangante that reacted in the titration.

2. n(Fe2+) in titration = 0.000484 x 5 = 0.00242 mol
Given the overall redox reaction in the titration, we can see that mole ratio of Fe2+: MnO4- = 5:1. So we multiply moles of manganate by 5 to get moles of Fe2+ in the titration.

3. n(Fe2+) total = 0.00242 x 10 = 0.0242 mol
We only worked out moles of Fe2+ in the titration, we need to know how many of moles of Fe2+ there was in total in the 250 cm3 solution so we multiply by 10 (as 250/25 = 10)

4. Mr (hydrated salt) = 6.3/0.0242 = 260.3305785123967
If there were 0.0242 mol of Fe2+ in total, then logically speaking there must have been 0.0242 mol of the hydrated salt (all the Fe2+ ions came from the hydrated salt and each hydrated salt contains 1 Fe2+ ion)

5. Mr (xH2O) = 260.3305785123967 - (56 + 32 + 64) = 108 (rounded to nearest whole number)
Now that we know the Mr of the entire hydrated salt, we will work out the total Mr of the water molecules in the hydrated salt. This must be a multiple of 18 as Mr of water is 18. To calculate this we take 260 and
subtract the Mr of Fe(56), S(32) and O in the sulphate(16 x 4 = 64)

6. x = 108/18 = 6
Now that we know the Mr of x water molecules is 108, and each water molecule has Mr 18 then there must have been 6 water molecules.
Hence x = 6
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