chemistry help (transition metal)

oxidation of iron with potassium nitrate in alkaline solution gives a red compound, R, which contains potassium, iron and oxygen only. R is a powerful oxidising agent and will oxidise iron(II) to iron(III).

25.0 cm^3 of a 0.100 mol dm^-3 solution of R oxidised 37.5 cm^3 of a solution of iron(II) contaning 55.6 g dm^-3 of FeSO(4).7H(2)O.

calculate the oxidation state of the iron in R.

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I don't even know where to begin. I guess writing equations would be a start, but I don't think I know any that will help me here.

Reply 1

Golden Maverick

4

mik1w

oxidation of iron with potassium nitrate in alkaline solution gives a red compound, R, which contains potassium, iron and oxygen only. R is a powerful oxidising agent and will oxidise iron(II) to iron(III).

25.0 cm^3 of a 0.100 mol dm^-3 solution of R oxidised 37.5 cm^3 of a solution of iron(II) contaning 55.6 g dm^-3 of FeSO(4).7H(2)O.

calculate the oxidation state of the iron in R.

---

I don't even know where to begin. I guess writing equations would be a start, but I don't think I know any that will help me here.

Right, molecular mass of

FeSO_4.7H_2O

:
(55.8 + 32 + 15 x 4) + (7 x 18) = 147.8 + 126

You need to find the mass of the

FeSO_4

so the ratio of this to total mass is:
147.8 / (147.8 + 126) = 0.52981...

Total mass:
55.6 x 37.5/1000 = 2.085g

Therefore mass without water:
0.52981... x 2.085 = 1.1255... g

Moles of

FeSO_4

:
1.1255... / 147.8 = 0.00761504... mol

Moles of solution:
0.1 x 25/1000 = 0.0025 mol

Divide one by the other:
0.0076150... / 0.0025 = 3.0460...

Therefore oxidation state of 3.

I think this is right, but can't be sure, anyone else?