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:

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.