Question

54 g of Aluminium and 320 g of oxygen gas react to form Aluminium oxide formed.Calculate  the mass of Al2O3 formed.

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Answer 1

Explanation:

4Al + 3O2 = 2Al2O3

Molar mass of Al = 27 g

Molar mass of Oxygen molecule = 32 g

Hence the Al and oxygen reacted in 1:1 molar ratio

Al is the limiting reactant as there is not sufficient of it to react with excess oxygen, as from the equation 1 mole of Al forms half mole of aluminium oxide, the molar mass of aluminium oxide is 27*2 + 16*3 = 102

Answer 2

The mass of Al₂O₃ produced is 102 g.

Given,

Mass of Aluminium = 54 g

Mass of Oxygen = 320 g

Aluminium and Oxygen react to form Aluminum Oxide

To Find,

Mass of Aluminum Oxide formed

Solution,

Aluminium and Oxygen react to form Aluminum Oxide

4Al + 3O₂ → 2Al₂O₃

So now we can say that 4 moles of Aluminium react with 3 moles of Oxygen to form 2 moles of Aluminium Oxide.

Mass of Aluminium = 54 g

Atomic Mass of Aluminum = 27

Moles = $\frac{Given mass}{Molar mass}$

Moles of Aluminum = $\frac{54}{27}$

Moles of Aluminum = 2 moles

Mass of Oxygen = 320 g

Molecular Mass of Oxygen = 32

Moles = $\frac{Given mass}{Molar mass}$

Moles of Oxygen = $\frac{320}{32}$

Moles of Oxygen = 10 moles

The moles of Aluminium are less as needed from stoichiometry, hence it acts as a limiting reagent.

Now the reaction would proceed as per the limiting reagent.

Using stoichiometry -

4 moles Al produce → 2 moles Al₂O₃

2 moles Al produce → 1 mole Al₂O₃

Moles of Al₂O₃ produced = Mass of 1 mole Al₂O₃

Moles of Al₂O₃ produced = 2(27) + 3(16)

Moles of Al₂O₃ produced = 54 + 48

Moles of Al₂O₃ produced = 102 g

Hence, the mass of Al₂O₃ produced is 102 g.

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