Question

Aluminium oxide is a white , crystalline powder . it can be used as an adsorbent or a catalyst for organic reactions and is formed when aluminium reacts with oxygen . A mixture of 74.38g of Aluminium and 56.27 g oxygen gas is allowed to react.
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1) write a balanced equation for the reaction above.
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2) identify the limiting reactanct by showing the calculation .
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3)determine the mass of the excess reactant present in the vessel when the reaction is complete.

Answer 1

(1): The balanced equation for the reaction of Al with O2 to form Al2O3 is

4Al(s) + 3O2(g) ---- > 2Al2O3(s)

(2): Given the mass of Al(s) = 74.38 g

Atomic mass of Al(s) = 27.0 g/mol

Hence moles of Al(s) = mass / atimic mass = 74.38 g / 27.0 g/mol =  2.755 mol

Given the mass of O2(g) = 56.27 g

molecular mass of O2(g) = 32.0 g/mol

Hence moles of O2(g) = mass / molecular mass = 56.27 g / 32.0 g/mol = 1.758 mol

4Al(s) + 3O2(g) ---- > 2Al2O3(s)

4 mol, --- 3 mol, ------- 2 mol

In the above balanced reaction, 4 mol of Al(s) reacts with 3 mol of O2(g) to form 2 mol of Al2O3(s).

Hence  2.755 mol of Al(s) that will react with the stoichiometric moles of O2(g)

= 2.755 mol Al x (3 mol O2 / 4 mol Al)

= 2.066 mol O2

Since the available moles of O2 (1.758 mol) is less than the required stoichiometric moles(2.066 mol), O2 acts as the limiting reactant and exhausts completely.(answer)

(3): Here the excess reactant is Al and O2 is the limiting reactant.

1.758 mol of O2 that will react with the moles of Al = (1.758 mol O2) x ( 4 mol Al / 3 mol O2) = 2.344 mol Al

Total moles of Al taken initially = 2.755 mol Al

Hence moles of Al remain unreacted = 2.755 - 2.344 = 0.411 mol Al

Hence mass of Al remain unreacted = (0.411 mol Al) x (27 g / mol Al) = 11.10 g (answer)