Chemical Calculations help!

I would highly appreciate if someone would explain in detail how to solve these. :)
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in an experiment, 6.08g of iron (ll) sulphate was heated until the decomposition was complete.
Calculate the mass of iron(ll)oxide formed and the total volume of gas produced measured at room temperature r.t.p

2FeSO4 (s) ---> Fe2O3 (s) + SO2 (g) + SO3 (g)
The mass of one mole of FeSO4 = 152g
The mass of one mole of Fe2O3 = 160g

i) calculate the number of moles of FeSO4 used: i got: no of moles= mass/1 mol mass=6.08/152=0.04 [correct?]

ii)use answer to part (i) to predict number of moles of FeO3 formed

iii) calculate the mass of iron(ll)xoide formed

iv) use answer to part (i) to predict number of moles of SO2 formed

v) what is the total volume of gas produced?

2FeSO4 (s) ---> Fe2O3 (s) + SO2 (g) + SO3 (g)

i) n(FeSO4) = m/MW
n(FeSO4) = 6.08/151.92
n(FeSO4) = 0.04002106372 ensure that you don't round off when calculating moles

ii) n(Fe2SO3) = n(FeSO4)/2 mole ratio 2:1
n(Fe2SO3) = 0.02001053186

iii) m(Fe2O3) = n x MW
m(Fe2O3) = 0.02001053186 x 159.7
m(Fe2O3) = 3.195681938g
m(Fe2O3) = 3.20g

iv) n(SO2) = 0.02001053186 mole ratio 1:1

v) v(SO2) = n x V
v(SO2) = 0.02001053186 x 24.79 reaction takes place at 25*C and using the data sheet - 24.79L
v(SO2) = 0.4960610848L
+
v(SO3) = 0.02001053186 x 24.79
v(SO3) = 0.4960610848L

Total volume of gas produced = 0.9921221696L
Therefore total volume = 0.99L

To avoid mistakes, ensure that you round off the values at the end of the calculation, when you have the answer. The round it off to the lowest significant figures from the question, which in this case it is 3. Therefore the values of the mass and volume are rounded to 3 sig figs.
Balancing the equation is also necessary as this provides the mole ratio which will help when calculating the mole of a reactant or product.