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A 37.2 g of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to Fe2O3....Asked by Anonymous
A 44.6 g of iron ore is treated as follows. The
iron in the sample is all converted by a series
of chemical reactions to Fe2O3. The mass of
Fe2O3 is measured to be 10.1 grams. What
was the percent iron in the sample of ore?
Answer in units of %
iron in the sample is all converted by a series
of chemical reactions to Fe2O3. The mass of
Fe2O3 is measured to be 10.1 grams. What
was the percent iron in the sample of ore?
Answer in units of %
Answers
Answered by
Damon
What fraction, F, of the mass of a Fe2O3 molecule is FE ?
total mass of molecule = 2(atomic mass Fe)+3(atomic mass O)
mass of Fe in that molecule = 2 (atomic mass Fe)
so
F = 2(atomic mass Fe)/[ the sum ]
then
Mass Fe = 10.1 * F
and percent = (Mass Fe/44.6)100
total mass of molecule = 2(atomic mass Fe)+3(atomic mass O)
mass of Fe in that molecule = 2 (atomic mass Fe)
so
F = 2(atomic mass Fe)/[ the sum ]
then
Mass Fe = 10.1 * F
and percent = (Mass Fe/44.6)100
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