4Fe(s) + 3O2(g) --> 2Fe2O3(s)
From the equation, we can see that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron(III) oxide. The molar mass of iron is 55.845 g/mol, and the molar mass of iron(III) oxide is 159.69 g/mol.
To determine the grams of iron needed to react completely with an excess of oxygen to form 40 g of iron(III) oxide, we can use stoichiometry. First, we need to calculate the number of moles of iron(III) oxide produced:
40 g Fe2O3 / 159.69 g/mol = 0.250 mol Fe2O3
According to the balanced chemical equation, 2 moles of iron(III) oxide are produced from 4 moles of iron. Therefore, the number of moles of iron needed is:
4 moles Fe / 2 moles Fe2O3 * 0.250 mol Fe2O3 = 0.500 mol Fe
Finally, we can convert the moles of iron to grams using the molar mass of iron:
0.500 mol Fe * 55.845 g/mol = 27.92 g Fe
Therefore, approximately 27.92 grams of iron will be needed to react completely with an excess of oxygen to form 40 grams of iron(III) oxide.
