Solve for x?

Question

here's a question i never seen b4:

Hydrated ammonium iron (II) sulfate crystals have the formula, (NH4)2SO4FeSO4xH2O. 8.325 g of the salt were dissolved in 250 mL of acidified water. A 25 mL portion of this solution was titrated with potassium manganate (VII) solution of concentration .148M. A volume of 2.25 mL was required. Calculate the value of x.

If you could answer this question, and kinda show the steps, that would be great.

Im self learning this stuff right now, so this really helps.

Thanks

Answer 1

if I can get the moles of the salt, I can then get the molar mass,  from that I can get the amount of hydration

notice that 1 mole  (NH4)2SO4FeSO4xH2O has 1 mole of Fe+2. it is the moles of Fe+2 that react with the KMnO4:

5Fe+2 (loses1 e- each) & 1 KMnO4 (takes 5 e-) -->

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moles of KMnO4:

0.00225 litres @ 0.148 mol/litre = 0.000333 moles KMnO4

0.000333 mol KMnO4 @ 5 mol Fe+2 /1 mol KMnO4 = 0.001665 moles Fe+2

@ 1:1 , 0.001665 moles Fe+2 = 0.001665 moles  (NH4)2SO4FeSO4xH2O

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molar mass of  (NH4)2SO4FeSO4xH2O:

8.325 g / 0.001665 mol = 5000 g/mol

well, that not realistic

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ok so we got a problem,  I don't think the volume of KMnO4 was 2.25 ml, maybe 22.25, 32.25, 42.25

whatever.  if the volume of KMnO4 was different, modify these calcs, or email me

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I am going to re-work the problem sort of guessing that the volume of KMnO4 should have been 28.25 ml

moles of KMnO4:

0.02825 litres @ 0.148 mol/litre = 0.004181 moles KMnO4

0.004181 mol KMnO4 @ 5 mol Fe+2 /1 mol KMnO4 = 0.02091 moles Fe+2

@ 1:1 , 0.02091 moles Fe+2 = 0.02091 moles  (NH4)2SO4FeSO4xH2O

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molar mass of  (NH4)2SO4FeSO4xH2O:

8.325 g / 0.02091 mol = 398 g/mol

without the water, (NH4)2SO4FeSO4 has a molar mass of 284,.... that leaves 114g for water. @ 18 g/mol ,... that,s enough for 6.3 H2O 's

so your formula would be

(NH4)2SO4FeSO4 dot 6H2O  (Mohr's salt)

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so check all of the data given , grams of unknwn, molarity of KMno4, & especially the mls