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21 May, 03:07

Aluminum metal reacts with hydrochloric acid to form aluminum chloride according to the reaction below. If 30.8 grams of aluminum react with 84 grams of HCl, how many grams of aluminum chloride can theoretically be formed? 2Al + 6HCl - -> 2AlCl3 + 3H2 You Answered

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  1. 21 May, 04:04
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    The amount in grams of aluminum chloride that can theoretically be formed from the 84 g of HCl and 30.8 g aluminium is 102.4 grams

    Explanation:

    Aluminum metal reacts with hydrochloric acid to form aluminum chloride according to the reaction below. If 30.8 grams of aluminum react with 84 grams of HCl, how many grams of aluminum chloride can theoretically be formed? 2Al + 6HCl - -> 2AlCl3 + 3H2

    Hydrochloric acid / Molar mass

    36.46 g/mol

    Aluminium chloride / Molar mass

    133.34 g/mol

    Aluminium / Atomic mass

    26.981539

    Hydrogen / Atomic mass

    1.00784 u

    Number of moles of aluminium = mass/molar mass = 30.8 g / 26.98 g/mol

    = 1.142 moles.

    Number of moles of HCl = mass/molar mass = 84/36.46 = 2.303 moles

    From the reaction, 2 moles of Al react with 6 moles of HCl, therefore, 1.142 moles of aluminium react with (1.142 * 6/2) moles or 3.425 moles of HCl. Since the calculated number of moles of required HCl is more than the available moles, then HCl is the limiting reagent.

    6 moles of HCl react with 2 moles of aluminium, then 2.303 moles will react with (2/6*2.303) or 0.76796 moles of aluminium to produce 0.76796 moles of aluminum chloride

    Therfore mass of aluminum chloride produced = (Number of moles of aluminium chloride produced) * (molar mass of aluminium chloride)

    0.76796 moles * 133.34 g/mol = 102.4 g
  2. 21 May, 05:20
    0
    Answer: The Theoretical yield of Aluminum Chloride = 152.19 g

    Explanation:

    In order to determine the theoretical yield of Aluminum Chloride we first determine the limiting reagent,

    Al = 27g/mol

    HCl = 1 + 35.5 = 36.5g/mol

    number of moles = mass/molar mass

    for Al:

    number of moles of Al = 30.8/27 = 1.14 moles

    number of moles of HCl = 84/36.5 = 2.30moles

    Hence, Al is the limiting reagent

    Therefore,

    since from the balanced reaction,

    2moles of Al react with HCl to give 2moles of Aluminum Chloride

    Therefore 1.14moles Al will give 1.14moles of Aluminum Chloride.

    molar mass of Aluminum Chloride = 27 + (35.5) * 3 = 133.5g/mol

    Theoretical yield of Aluminum Chloride = number of moles x molar mass = 1.14 x 133.5 = 152.19 g
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