Question

For the following reaction, 46.0 grams of iron are allowed to react with 18.4 grams of oxygen gas .

iron(s) + oxygen(g) ---> iron(III) oxide(s)

a. What is the maximum mass of iron(III) oxide that can be formed?

b. What is the formula of the limiting reagent?

c. What mass of the excess reagent remains after the reaction is complete?

Answer 1

The balanced equation is

4 Fe (s) + 3 O₂ (g) -----------> 2 Fe₂O₃ (s)

Number of moles of Fe = mass/mol.wt. =46.0 g/( 55.845 g/mol) = 0.823708 mol

Number of moles of O₂ = mass/mol.wt. = 18.4 g/( 32.00 g/mol) = 0.575 mol

According to balanced equation,

3 moles of O₂ reacts with 4 moles of Fe

So, 0.575 mol of O₂ would react with 0.575 mol * 4/3 = 0.766667 mol of Fe

But, the number of moles of Fe present is more than required.

That is

Fe is the excess reagent and O₂ is the Limiting reagent.

According to balanced equation,

3 moles of O₂ produces 2 moles of Fe₂O₃

So, 0.575 mol of O₂ would produce 0.575 mol * 2/3 = 0.383333 mol of Fe₂O₃

Mass of Fe₂O₃ = moles * mol.wt. = 0.383333 mol * 159.69 g/mol = 61.2145 g

Mass of Fe₂O₃ = 61.2 g

According to balanced equation,

3 moles of O₂ reacts with 4 moles of Fe

So, 0.575 mol of O₂ would react with 0.575 mol * 4/3 = 0.766667 mol of Fe

But, the number of moles of Fe present is more than required.

That is

Fe is the excess reagent

Number of moles of Fe present in excess = 0.823708 – 0.766667 = 0.057042 mol

Mass of Fe remains after the reaction = moles * mol.wt. = 0.057042 mol * 55.845 g/mol = 3.19 g

Answer :

Mass of Fe₂O₃ = 61.2 g

O₂ is the Limiting reagent

Mass of excess reagent (Fe) remains after the reaction = 3.19 g