4.80 g of oxygen is used to burn 0.150 mol of iron. what mass Fe2O3 will be produced? what mass of Fe will be left over at the end of the reaction? what mass of O2 will be left over at the end of the reaction?
Answers
4.80 g of oxygen is used to burn 0.150 mol of iron. what mass Fe2O3 will be produced? what mass of Fe will be left over at the end of the reaction? what mass of O2 will be left over at the end of the reaction?
Explanation:
4 Fe + 3 O2 → 2 Fe2O3
(4.80 g O2) / (31.99886 g O₂/mol) = 0.1500 mol O₂
0.150 mole of Iron(Fe) would react completely with 0.150 x (3/4) = 0.1125 mole of O₂, but there is more O₂ present than that, hence O₂ is in excess and Fe is the limiting reactant.
(0.150 mol Fe) x (2/4) x (159.6887 g Fe₂O3/mol) = 12.0 g Fe₂O3
(0.1500 mol O₂ initially) - (0.1125 mol O₂ reacted) = 0.0375 mol O₂ left over
Answer:
Hey bro ur ans is absolutely right but I would like to add one more point to it that after the reaction has taken place 0 gm of Fe will remain as it's asked in the question . Hope u understand