A quantity of 10.0 g Iron is reacted with 20.0 g Oxygen gas. Calculate the the amount of Iron(III) oxide that could be made. 4Fe(s) + 3O2(g)  2Fe2O3(s)

Answer 1

First, write the correctly balanced equation

4Fe(s) + 3O₂(g) ==> 2Fe₂O₃(s) ... balanced equation

Next, since we are given the quantity of BOTH reactants, we must find which one is limiting. An easy way to do this is to simply divided the MOLES of each reactant by the corresponding coefficient in the balanced equation, and whichever value is less represents the limiting reactant.

For Fe we have: 10.0g Fe x 1 mol Fe / 55.85 g = 0.1791 moles Fe (÷4->0.045)

For O₂ we have: 20.0 g O₂ x 1 mol O₂ / 32 g = 0.625 moles O₂ (÷3->0.21)

Since 0.045 is less than 0.21, Fe is the limiting reactant, and the MOLES of Fe (0.1792 mols) will determine

how much product can be made.

Finally, we use the mols of Fe and the stoichiometry of the balanced equation to calculate mols of Fe₂O₃ that can theoretically be produced

0.1792 mol Fe x 2 mol Fe₂O₃ / 4 mol Fe = 0.0896 mols Fe₂O₃ (3 sig. figs.)

If you want the mass of Fe₂O₃, then use the molar mass of Fe₂O₃ to convert:

0.0896 mol Fe₂O₃ x 159.7 g / mol = 14.3 g Fe₂O₃ (3 sig. figs.)