What weight of sample containing 8.00 % Fe3O4 must be taken to obtain a precipitate of Fe(OH)3 that, when ignited to Fe2O3, weighs 150 mg?

3 answers

Work it backwards.
You want to end up with 150 mg Fe2O3. How much Fe(OH)3 must you have?
150 mg Fe2O3 x [2*molar mass Fe(OH)3/molar mass Fe2O3] = ? mg Fe(OH)3
What must be the mass Fe3O4?
?mg Fe(OH)3 x [2*molar mass Fe3O4/3*molar mass Fe2O3] = ? mg Fe3O4.
Since that mg Fe3O4 is only 8%, then
mass sample x 0.08 = ?mg Fe3O4 and solve for mass sample.
Those factors I used come from this.
2Fe3O4 ==> 3Fe2O3
Fe2O3 ==> 2Fe(OH)3
I didn't balance the oxygen atoms since we are working only with the Fe

Check my work.

Thanks a lot

The answer is 2.43g :))