In a reaction, 82.00g of sodium reacts with 74.00g of ferric oxide [iron(iii)oxide] to from sodium oxide and iron metal. calculate the mass of solid iron produced.

The reaction between Na and Fe₂O₃ is as follows;
6Na + Fe₂O₃ --> 3Na₂O + 2Fe
stoichiometry of Na to Fe₂O₃ is 6:1
the number of Na moles present - 82.00 g /23 g/mol = 3.56 mol
number of Fe₂O₃ moles present - 74.00 g / 159.7 g/mol = 0.46 mol 
Limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed depends on the amount of limiting reactant present.
If Na is the limiting reactant, number of  Fe₂O₃ moles required -
3.56/6 = 0.59 mol 
however only 0.46 mol of Fe₂O₃ are present. Therefore Fe₂O₃ is the limiting reactant. 
Stoichiometry of Fe₂O₃ to Fe is 1:2
number of Fe moles formed - 0.46 mol x 2 = 0.92 mol 
the mass of Fe formed = 0.92 mol x 55.8 g/mol = 51.33 g of Fe is formed