Question

In the following chemical reaction, how many grams of the excess reactant are left over when 0.45 moles of Al react with 1.25 moles of Cl2 ?
2Al + 3Cl2 → 2AlCl3

Answer 1

Hi,

from the balanced reaction, we can interpret,

2 moles of Al react with= 3 moles of Cl2

1 mole of Al reacts with= 3/2 moles of Cl2

0.45 moles of Al reacts with= (3/2)*0.45 moles of Cl2

= 0.675 moles of Cl2

clearly Cl2 is in excess.

therefore the moles of Cl2 left after reaction= 1.25- 0.675 moles= 0.575 moles of Cl2

Molecular weight of Cl2= 35.5*2= 71 g/gmol

grams of Cl2 left= number of moles of Cl2 left* molecular weight of Cl2

= 0.575*71= 40.825 g of Cl2 left after the reaction.