Question

In the reaction of formation of sodium chloride (NaCl), 15 g of sodium is placed in contact with 20 g of chlorine. What is the reactant in excess and
its mass in excess?

Answer 1

2Na(s) + Cl2(g) => 2NaCl(s)
Convert given data to moles and divide each by the respective coefficient in the balanced equation. The smaller value is the limiting reagent and the larger value is the one that will remain in excess.

moles Na = (15g/23g-mol^-1)mole = 0.652 mole Na

moles Cl2 = (20g/70g-mol^-1)mole = 0.286 mole Cl2

Na => (0.652/2)=0.326
Cl2 => (0.286/1)= 0.286
Cl2 is the Limiting reagent and Na will remain in excess upon completion of the reactions.

Answer 2

Thanks! =)