Question

2.
0.5 g Fe3+ is precipitated as hydrated ferric oxide. During ignition, 90% of iron is converted into
Fe203 and the remainder is present as Fe3O4
(a) What does the ignited precipitate weigh?
(b) What should it would have weighed if all the iron were in Fe2o3 form?​

Answer 1

Answer:

a) 0.711g

b)0.685g

Explanation:

a)

90% of 0.5g Fe³⁺ = 0.45g of Fe become Iron(III) oxide ie..Fe₂O₃

now in Fe₂O₃→ 2 moles of Fe are present per 3 moles of O thus 3/2 = 1.5 moles of O present per mole of Fe

in 0.45g Fe³⁺ 0.45/56 moles of Fe are present which means (0.45/56×1.5) = 0.012053571 moles of oxygen are added (mole=mass/relative molecular mass), then the total mass is 0.45+(0.012053571×16)=0.642g

then, the remaining 10% becomes Iron(II,III) oxide ie..Fe₃O₄ this is = 0.05g

in which 3 moles of Fe are present per 4 moles of O thus 4/3 = 1.33 moles of O present per mole of Fe

in 0.05g Fe³⁺ 0.05/56 moles of Fe are present which means (0.05/56×1.33) = 0.001190476 moles of oxygen are added, then the total mass is 0.05+( 0.001190476×16)=0.069048g

overall mass = 0.71104g

b) then 100% of 0.5g  = 0.5g

using the previously described method we have total mass = [(0.5/56×1.5)×16]g + 0.5g = 0.685g