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9 October, 06:13

When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:

2 Al (s) + 3 Cl2 (g) - -> 2 AlCl3 (s)

b) How many moles of AlCl3 are formed?

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Answers (2)
  1. 9 October, 08:37
    0
    1.5 mole

    Explanation:

    Step 1:

    The balanced equation for the reaction. This is illustrated below:

    2Al (s) + 3Cl2 (g) - -> 2AlCl3 (s)

    Step 2:

    Determination of the masses of Al and Cl2 that reacted from the balanced equation. This is illustrated below:

    Molar Mass of Al = 27g/mol

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

    Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

    Mass of Cl2 from the balanced equation = 3 x 71 = 213g

    From the balanced equation,

    54g of Al reacted.

    213g of Cl2 reacted

    Step 3:

    Determination of the limiting reactant.

    This is illustrated below:

    From the balanced equation above,

    54g of Al reacted with 213g of Cl2.

    Therefore, 40.5g of Al will react with = (40.5 x 213) / 54 = 159.75g of Cl2.

    From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

    Step 4:

    Determination of the number of mole in 40.5g of Al. This is illustrated below:

    Molar Mass of Al = 27g/mol

    Mass of Al = 40.5g

    Number of mole of Al = ?

    Number of mole = Mass/Molar Mass

    Number of mole of Al = 40.5/27

    Number of mole of Al = 1.5 mole

    Step 5:

    Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

    2Al (s) + 3Cl2 (g) - -> 2AlCl3 (s)

    From the balanced equation above,

    2 moles of Al produced 2 moles of AlCl3.

    Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

    From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.
  2. 9 October, 09:31
    0
    1.5 moles AlCl3 will be formed

    Explanation:

    Step 1: Data given

    MAss of Al = 40.5 grams

    Mass of Cl2 = 212.7 grams

    Atomic mass Al = 26.98 g/mol

    Molar mass Cl2 = 70.9 g/mol

    Step 2: The balanced equation

    2 Al (s) + 3 Cl2 (g) - -> 2 AlCl3 (s)

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles Al = 40.5 grams / 26.98 g/mol

    Moles Al = 1.50 moles

    Moles Cl2 = 212.7 grams / 70.9 g/mol

    Moles Cl2 = 3.0 moles

    Step 4: Calculate limiting reactant

    For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

    Al is the limiting reactant. It will completely be consumed (1.5 moles).

    Cl2 is in excess. There will react 3/2 * 1.5 = 2.25 moles Cl2

    There will remain 3.0 - 2.25 = 0.75 moles Cl2

    Step 5: Calculate moles AlCl3

    For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

    For 1.5 moles AlC we'll have 1.5 moles AlCl3

    1.5 moles AlCl3 will be formed
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