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14 April, 03:44

If 5.0 of al react with 9.0 g of O2, how many grams of Al2O3 (101.56 g/mol) can be formed

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  1. 14 April, 04:16
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    9.394 g of Al₂O₃

    Explanation:

    The reaction between Aluminium and Oxygen is given by the balanced equation;

    4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

    We are given;

    Mass of Aluminium as 5.0 g Mass of Oxygen as 9.0 g

    We are required to determine the mass of Al₂O₃ formed.

    We are going to use the following simple steps;

    Step 1: Calculate the number of moles of Aluminium and Oxygen

    Moles = Mass : Molar mass

    Moles of Al;

    Molar mass of Al = 26.98 g/mol

    Moles of Al = 5.0 g : 26.98 g/mol

    = 0.185 moles

    Moles of Oxygen gas

    Molar mass of O₂ = 32.0 g/mol

    Moles O₂ = 9.0 g : 32.0 g/mol

    = 0.28125 moles

    From the equation, four moles of Aluminium reacts with three moles of oxygen gas. Therefore, Aluminium is the rate limiting reagent while oxygen gas is in excess. Step 2: Calculate the number of moles of Al₂O₃ formed

    4 moles of Al reacts to produce 2 moles of Al₂O₃

    Thus, the mole ratio of Al to Al₂O₃ is 2 : 1

    Thus, moles of Al₂O₃ = Moles of Al : 2

    = 0.185 moles : 2

    = 0.0925 moles

    Step 3: Calculate the mass of Al₂O₃ formed

    Mass = Moles * Molar mass

    Molar mass of Al₂O₃ is 101.56 g/mol

    = 0.0925 moles * 101.56 g/mol

    = 9.394 g

    Therefore, the mass of Al₂O₃ formed is 9.394 g
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