04 mol = 24.0*0.04 = 0.96 L gas produced.......
In this case 2FeSO4 (s) ---> Fe2O3 (s) + SO2 (g) + SO3 (g)
The mole ratio is 2 : 1 :1 : 1
So to calculate the moles of Fe2O3 divide the moles of FeSO4 by two to calculate the moles of FeO3.
Hopes this helps.
i) calculate the number of moles of FeSO4 used: i got: no of moles= mass/1 mol mass=6.08/152=0.04 [correct?] OK - correct.
ii)use answer to part (i) to predict number of moles of Fe2O3 formed
From the equation:
2mol FeSO4 produces 1 mol Fe2O30.04 mol FeSO4 will produce 0.02mol Fe2O3
iii) calculate the mass of iron(ll)xoide formed
0.02*160 = 3.2g formed
iv) use answer to part (i) to predict number of moles of SO2 formed
2mol FeSO4 produce 1 mol SO2
0.04mol FeSO4 will produce 0.02 mol SO2
v) what is the total volume of gas produced?
Total mol of gas = 0.02mol SO2 + 0.02mol SO3 = 0.04 mol in total
At RTP, 1 mol = 24.0L
0.04 mol = 24.0*0.04 = 0.96 L gas produced.
