Question

The reaction between solid sodium and iron(iii) oxide is one in a series of reactions that inflates an airbag. 6na(s) + fe2o3(s) → 3na2o(s) + 2fe(s) if 100.0 g na and 100.0 g fe2o3 are used in this reaction, determine:

Answer 1

Answers:

a. the limiting reagent.
Fe₂O₃ is limiting reagent

b. the reactant in excess.
Na is in excess.

c. the mass of solid iron produced.
69.93 g of Fe

d. the mass of excess reactant that remains after the reaction is complete.
13.66 g of Na is left unreacted.

Solutions:

The reaction is as,,

6 Na + Fe₂O₃ → 3 Na₂O + 2 Fe

According to equation,

137.88 g (6 mole) Na reacts with = 159.69 g (1 mole) of Fe₂O₃
So,
100 g of Na will require = X g of Fe₂O₃

Solving for X,
X = (100 g × 159.69 g) ÷ 137.88 g

X = 115.81 g of Fe₂O₃

But we are provided with only 100 g of Fe₂O₃, it means Fe₂O₃ is Limiting reagent and Na is in Excess. Therefore, Fe₂O₃ will control the amount of product formed.
So, as,

159.69 g (1 mole) Fe₂O₃ produced = 111.68 g (2 mole) of Fe
So,
100 g of Fe₂O₃ will produce = X g of Fe

Solving for X,
X = (100 g × 111.68 g) ÷ 159.69 g

X = 69.93 g of Fe
As,
159.69 g (1 mole) Fe₂O₃ required = 137.88 g of Na
So,
100 g of Fe₂O₃ will require = X g of Na

Solving for X,
X = (100 g × 137.88 g) ÷ 159.69 g

X = 86.34 g of Na

So, the Na left behind after reaction is 100 - 86.34 = 13.66 g.