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10 March, 22:15

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of each substance are present after 72.5 g of aluminum nitrite and 58.6 g of ammonium chloride react completely?

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  1. 10 March, 22:24
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    Given dа ta:

    The mass of aluminium nitrite is 72.5 g

    The mass of ammonium chloride is 58.6 g

    The balanced chemical equation for the reaction is given as follows.

    Al (NO2) 3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O

    The number of moles can be determined by the formula given as follows.

    Number of moles = Mass / Molar mass

    The molar mass of aluminum nitrate and ammonium chloride is 164.998 g/mol and 53.49 g/mol respectively.

    inserting the respective values in the formula given above.

    Moles of Al (NO2) 3 = 72.5 g / 164.998 g/mol = 0.439 mol

    Moles of NH4Cl = 58.6 g / 53.49 g/mol = 1.096 mol

    From the balanced equation,

    3 moles of ammonium chloride requires 1 mole of aluminum nitrate.

    So, 1 mole of ammonium chloride requires 1 / 3 mole of aluminum nitrate.

    Thus, 1.096 mole of ammonium chloride will require (1 / 3) * 1.096 = 0.3653 mole of aluminum nitrite.

    Here, the amount of aluminum nitrate is more than the required amount so ammonium chloride is the limiting reagent.

    From the balanced chemical equation, 3 mole of ammonium chloride gives 1 mole of aluminum chloride.

    So, 1.096 mole of ammonium chloride will give;

    3 = 1

    1.096 = x

    x = (1.096 * 1) / 3 = 0.3653 mole of aluminum chloride.

    Therefore, the number of moles of aluminum chloride is 0.3653 mol.

    Since the molar mass of aluminum chloride is 133.34 g/mol

    Substitute the respective values in the formula given above.

    0.3653 mol = Mass / 133.34 g/mol

    Mass = 0.3653 mol * 133.34 g/mol = 48.71 g

    Therefore, the mass of aluminum chloride produces is 48.71 g.

    The Ammonium chloride is completely used up in the reaction.

    The amount of alumium nitrite used is = Number of moles * Molar mass = 0.3653 * 164.998 = 60.27

    Mass of alminium nitrite left = 72.5 - 60.27 = 12.23g
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