Aluminium reacts with oxygen gas to form aluminium oxide ( Al ₂ O ₃ ). If 54 ~g of aluminium is reacted with 64 ~g of oxygen gas, the maximum mass (in g) of Al ₂ O ₃ that can be formed is : [Given molar masses in g mol ⁻¹: Al =27, O =16 ]

Step-by-step solution

The balanced chemical equation for the reaction is: 4 Al + 3 O ₂ 2 Al ₂ O ₃ Moles of Al initially present = 54 27 = 2 ~mol Moles of O ₂ initially present = 64 32 = 2 ~mol To find the limiting reagent, we compare the required molar ratio with the available molar ratio. According to the balanced equation, 4 ~mol of Al require 3 ~mol of O ₂ . So, 2 ~mol of Al will require 3 4 2 = 1.5 ~mol of O ₂ . Since 2 ~mol of O ₂ are available, O ₂ is in excess and Al is the limiting reagent. The amount of product formed depends e