A 36.5 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 grams. What was
the percent iron in the sample of ore?
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To find the percent of iron in the sample of ore, we need to compare the mass of the iron in the sample to the mass of the entire sample.
First, let's calculate the mass of iron in the sample. We know that all of the iron in the sample is converted to Fe2O3. The molar mass of iron (Fe) is 55.85 g/mol and the molar mass of Fe2O3 is 159.69 g/mol.
Since the sample contains 36.5 g of iron ore and all of it is converted to Fe2O3, we can calculate the amount of iron in the sample using the equation:
mass of iron = (mass of Fe2O3 / molar mass of Fe2O3) x molar mass of iron
mass of iron = (10 g / 159.69 g/mol) x 55.85 g/mol
mass of iron = 3.465 g
Next, let's calculate the percent of iron in the sample by dividing the mass of iron by the mass of the entire sample and then multiplying by 100:
percent iron = (mass of iron / mass of sample) x 100
percent iron = (3.465 g / 36.5 g) x 100
percent iron = 9.5%
Therefore, the percent of iron in the sample of ore is 9.5%.