posted by Sarah .
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?
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