Question

How many grams of iron could be

recovered from 80.0 g of an ore that is known to be 76.0 % Fe3O4?

Answer 1

43.8 g of iron can be recovered

Step-by-step explanation:

First get the mass of Fe3O4 in the ore:

80g Ore * (76 g Fe3O4 / 100 Ore) = 60.8 g Fe3O4

Then get number of moles of Fe3O4:

60.8 g Fe3O4 * 1 mol / 231.535 g = 0.262 mol Fe3O4

**231.535 g is the molar mass of Fe3O4

Then get number of moles of Fe:

0.262 mol Fe3O4 * (3 Fe / 1 Fe3O4) = 0.786 mol Fe

** 3 atoms of Fe in every Fe3O4

Finally convert moles of Fe to grams of Fe:

0.786 mol * 55.845 g/mol = 43.79 g Fe --round to 3 sig figs--> 43.8 g Fe