From a 45g sample of an iron ore containing Fe3O4, 1.56g of Fe is obtained by:Fe3O4 + 2C --gt; 3Fe + 2CO2What is the percent of Fe3O4 in the ore?Clear explanations please! Thanks!
Ok, the standard atomic weight of iron is 55.8 g/mol, oxygen is 16, and carbon is 12. First we find the mol of iron obtained from the reaction that is by dividing the mass of iron obtained with the standard atomic weight of iron, 1.56/55.8, we will get 0.03 mol. Therefore 0.01 mol of Fe3O4 and 0.02 mol of C will produce 0.03 mol of iron and 0.02 mol of CO2. Fe3O4 + 2C -- 3Fe + 2CO2 0.01Fe3O4 + 0.02C -- 0.03Fe + 0.02CO2 Then we find the mass of Fe3O4, by multiplying the reactant mol with the ionic mass, 0.01 x 231.4 2.314 g. Therefore only 2.314 g of Fe3O4 is used in this reaction from 45 g sample of iron ore, the percentage is 5.14%.
