chem! (40 points)reacting mass

1a) When iron(lll) hydroxide was heated to a high temperature, iron(lll) oxide was formed.
2Fe(OH)3 → Fe2O3 + 3H2O
Calculate the maximum mass of water formed from 5.35 g of iron(lll) hydroxide.
b) Another oxide of iron consist of 72.4% iron by mass. Calculate the empirical formula of this oxide.
(Relative atomic massws: H = 1.0, O = 16.0, Fe = 55.8)

a)
2Fe(OH)3 → Fe2O3 + 3H2O
Mole ratio Fe(OH)3 : H2O = 2 : 3

Molar mass of Fe(OH)3 = 55.8 + 3(16 + 1) = 106.8 g /mol
No. of moles of Fe(OH)3­ used = 5.35/106.8 = 0.05009 mol
Maximum no. of moles of H2­O formed = 0.05009 x (3/2) = 0.07514 mol
Molar mass of H2O = 1x2 + 16 = 18 g /mol
Maximum mass of H2O formed = 0.07514 x 18 = 1.35 g

b)
Mole ratio Fe : O
= 72.4/55.8 : (100 - 72.4)/16
= 1.30 : 1.73
= 1 : 1.33
= 3 : 4

Empirical formula = Fe3O4