Mining companies use this reaction to obtain iron from iron ore: fe2o3(s)+3 co(g)→2 fe(s)+3 co2(g) the reaction of 167 g fe2o3 with 83.5 gco produces 73.1 gfe. part a determine the limiting reactant.

1) To determine the molar mass you have to consult the periodic table and see the atomic mass of each element. If the element is repeted, for example 3 times, you multiple the atomic mass per 3.
CO= 12 + 16= 28 g/mol
Fe₂O₃= 19 x2 + 16x 3= 159,69 g/mol  

2) to calculate the limiting reactant you need to pass grams to moles.
 moles is calculated by dividing mass by molar mass

mass of Fe₂O₃ = 167 g
mass of CO = 83.5  g
molar mass of Fe₂O₃ = 159,69 g/mol  
molar mass of CO= 28,01 g/mol

moles of Fe₂O₃ = 167/ 159,69 = 1,046 mol of Fe₂O₃
moles CO= 83.5/ 28,01 = 2.981 mol of CO
 
               Fe2O3(s) + 3 CO(g) = 2 Fe(s) + 3 CO2(g)
3)By looking at the balanced equation, you can see that 1 mol of Fe₂O₃ needs 3 moles of CO to fully react . So, lets choose a random reactant for example, Fe₂O₃. To react   1,046 moles of Fe₂O₃, you need 3x1,046 moles of CO.
CO needed = 3x 1,046 moles= 3,138 mol of CO

You have less that 3,138 moles, you only have  2.981 mol of CO to react. So this means the limiting reagent is the CO. (The molecule that has molecules that are left is never the limiting reactant).

Do u have the n? or maybe the m?