When 60.0g of aluminum reacts with hydrochloric acid, how many grams of aluminum chloride are prepared? Balance the reaction first and

Question

When 60.0g of aluminum reacts with hydrochloric acid, how many grams of
aluminum chloride are prepared? Balance the reaction first and then solve the problem

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Sigridomena 4 years 2021-07-27T00:00:18+00:00 1 Answers 93 views 0

Answers ( )

  1. haidang
    0
    2021-07-27T00:01:26+00:00
    Reply

    Answer:

    296.67 g of AlCl₃.

    Explanation:

    We’ll begin by writing the balanced equation for the reaction. This is illustrated below:

    2Al + 6HCl —>2AlCl₃ + 3H₂

    Next, we shall determine the mass of Al that reacted and the mass of AlCl₃ produced from the balanced equation. This can be obtained as follow:

    Molar mass of Al = 27 g/mol

    Mass of Al from the balanced equation = 2 × 27 = 54 g

    Molar mass of AlCl₃ = 27 + (35.5×3)

    = 27 + 106.5

    = 133.5 g/mol

    Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g

    SUMMARY:

    From the balanced equation above,

    54 g of Al reacted to produce 267 g of AlCl₃.

    Finally, we shall determine the mass of AlCl₃ produced by the reaction of 60 g of Al. This can be obtained as follow:

    From the balanced equation above,

    54 g of Al reacted to produce 267 g of AlCl₃.

    Therefore, 60 g of Al will react to produce = (60 × 267)/54 = 296.67 g of AlCl₃

    Thus, 296.67 g of AlCl₃ were obtained from the reaction.

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