Question

If 112 g of fe combines with 48 g of o2 how much fe2o3 is formed?

Answer 1

Answer:

48 g of O2 gives 216g of FeO.

Explanation:

Amount of iron oxide formed when 112 g of Fe reacts with 48g of O2 is 108 grams.

Reaction of above problem is

2Fe+O2------->2FeO

So 2 moles of iron react with 1 mole of oxygen to give 2 moles of FeO.

That is 112 g of iron reacts with 32 g of oxygen but since oxygen is 48 grams as given hence oxygen is limiting reagent.

●A limiting reagent is one which is present in limited amount and the concentration of product formed depends on limiting reagent.

●So here amount of iron oxide formed depends on amount of oxygen.

●According to equation we have

32g of oxygen gives 2×(56+16)g of FeO

32 g of O2 gives 144 grams of FeO

1g of O2 gives 144/32 g of FeO

So,48 g of O2 gives (144/32)×48g

48 g of O2 gives 216g of FeO.