Question

A mixture of 82.49 g of aluminum and 117.6 of oxygen is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present when the reaction is complete.

Answer 1

The balanced equation is

4Al + 3O₂ -----> 2 Al₂O₃

Number of moles of Al = 82.49g / 27.0 g/mol = 3.055 mole

Number of moles of O2 = 117.6g / 32.0 g/mol = 3.675 mole

from the balanced equation we can say that

4 mole of Al requires 3 mole of O2 so

3.055 mole of Al will require

= 3.055 mole of Al *(3 mole of O2 / 4 mole of Al )

= 2.29 mole of O2 is required

But we have 3.675 mole of O2 which is in excess so O2 is excess reactant and aluminium is limiting reactant

number of moles of excess reactant = 3.675 - 2.29 = 1.385 mole of O2

1 mole of O2 =32.0 g

1.385 mole of O2 = 44.3 g

Therefore, the mass of excess reactant is 44.3 g