Solid iron metal reacts with oxygen gas to give iron (III) oxide according to the following unbalanced chemical equation: Fe (s) + O2 (g) ® Fe2O3 (s) If you react 42.88 g of iron with excess O2, how many grams of iron (III) oxide should form? I have balanced the equation and have 4 Fe + 3 O2 ==> 2Fe2O3
I also have calculated the molar mass of iron and oxygen..
Fe= 4 x 55.85 = 223.4 g/mol
O = 6 x 16.00 = 96 g/mol
total = 223.4 +96 = 314.4 g/mol

I then took the initial # provided in the equation and divided it by mass number of Fe (42.88 g/mol / 223.4 g/mol) multiplied by the coefficients 6mol O2 divided by coefficient of 4 mol Fe and finally divided by total molar mass of Fe2O3 319.4 g/mol.
I got a very big number (877.9g/mol Fe2O3) so I feel like I didn't solve it correctly, can someone check for me please?