What is the minimum mass of magnetite Fe3O4, a iron ore, from which 7.5kg of pure iron could be extracted?

Question

What is the minimum mass of magnetite Fe3O4, a iron ore, from which 7.5kg of pure iron could be extracted?

Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.

Answer 1

Hey Yasmine!

First, we have to start off with finding out how much Fe is in Fe₃O₄ by percentage composition. We can do this by finding out how taking the mass of each individual element in Fe₃O₄ and then dividing by the mass of Fe₃O_4.

To save time, Fe composes 72.36% and O composes 27.64% of Fe₃O₄. The question asks for mass of Fe, so we can ignore the percent composition of the Oxygen. To find the minimum mass needed to extract 7.5 kg (or 7500 grams) of Fe, we need to express the 7.5kg mathematically. We know that some amount of Fe₃O₄ with a percent composition of 72.36% Fe will yield 7.5kg of Fe. Therefore,

(.7236) (x) =7.5kg

Solving for x, we get that we need 10.36kg of Fe₃O₄ in order to get 7.5kg of Fe out of it.

Hope this helped!