Question

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants. 4Al(s)+3O2(g)→2Al2O3(s)

16 mol Al, 13 mol O2

11.4 mol Al, 9.5 mol O2

4 mol Al, 2.6 mol O2

1 mol Al, 1 mol O2

Can you please explain how to figure this out?

Answer 1

4Al(s)+3O2(g)→2Al2O3(s)

from balanced equation

3 moles of O2 react with 4 moles of Al

13 moles of O2 react with = 4*13/3   = 17.34 moles of Al is required.

Al is limitinf reactant

2.

from balanced equation

3 moles of O2 react with 4 moles of Al

9.5 moles of O2 react with = 4*9.5/3   = 12.67 moles of Al is required.

Al is limiting reactant

3

from balanced equation

4 moles of Al react with 3 moles of O2

4 moles of Al react with 3 moles of O2 is required

   O2 is limitinf reactant

4.

3 moles of O2 react with 4 moles of Al

1 moles of O2 react with = 4*1/3   = 1.33 moles of Al is required.

Al is limiting reactant