One of the reactions for rusting iron is as follows: 4fe + 3o2 → 2fe2o3 (mm fe: 55.85 g/mol; mm o2=32 g/mol; mm fe2o3=159.70 g/mol) if 63.98 g of oxygen gas is completely consumed, how many moles of iron (iii) oxide are formed? a. 1.333 mol b. 3071 mol c. 2.999 mol d. 6812 mol

he balanced equation for the reaction is
4Fe + 3O₂ ---> 2Fe₂O₃ 
stoichiometry of O₂ to Fe₂O₃  is 3:2
                   
number of O₂ moles used up - mass present / molar mass of O₂
number of O₂ moles = 63.98 g / 32 g/mol = 1.999 mol
 
if 3 mol of O₂ forms 2 mol of Fe₂O₃  
then 1.999 mol of O₂ forms - 2/3 x 1.999 mol = 1.333 mol
answer is A) 1.333 mol of Fe₂O₃  are formed

Ok where is the image ?