A 43.6 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 11.5 grams. What was the percent iron in the sample of ore? Answer in units of %.

Question

Alannah Solomon

Chemistry

3 October, 18:24

A 43.6 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 11.5 grams. What was the percent iron in the sample of ore? Answer in units of %.

+1

Answers (1)

Know the Answer?

Branson Ewing · Answer 1

The mass of Fe₂O₃ contains both iron and oxygen. First, we determine the amount of iron present in this compound. Then, because the mass of iron is conserved, the amount of iron present in the original ore will be the same.

Percentage iron in Fe₂O₃ = (2 * 56) * 100 / (2 * 56 + 3 * 16) = 70%

Therefore, out of the 11.5 grams of Fe₂O₃,

11.5 * 0.7 = 8.05 grams is iron

In the original ore, the percentage of iron is given by:

(8.05 * 100) / 43.6

The original ore had 18.5% iron.