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23 June, 15:41

If 5 moles of aluminum Al was reacted with 10 moles bromine Br2, all five moles of aluminum would react, with 7.5 moles bromine. (2:3 mole ratio)

Now assume 3 moles Al and 4 moles Br2 react

a) which chemical is the limiting reactant?

b) what chemical must be the excess reactant

c) How much (in moles) AlBr3 gets produced?

d) If all the limiting reactant gets used up, how much of the excess reactant is left?

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  1. 23 June, 15:51
    0
    Q1)

    we have been told that molar ratio of Al to Br₂ is in the 2:3 ratio

    the balanced equation for the reaction between Al and Br₂ is as follows;

    2Al + 3Br₂ - - > 2AlBr₃

    therefore the molar ratio of Al to Br₂ is 2:3 as stated previously

    when 3 mol of Al react with 4 mol of Br₂

    we have to find the limiting reactant. Limiting reactant is the reagent that is fully consumed during the reaction.

    If we assumed Al to be the limiting reactant

    if 2 mol of Al react with 3 mol of Br₂

    then 3 mol of Al would react with - 3/2 x 3 = 4.5 mol of Br₂

    however only 4 mol of Br₃ is present, this means that Br₂ is the limiting reactant.

    Answer for limiting reactant is Br₂

    Q2)

    excess reactant is the reagent in which only a fraction of the amount present is used up in the reaction

    when Br₂ is the limiting reactant

    if 3 mol of Br₂ react with 2 mol of Al

    and 4 mol of Br₂ present

    then 4 mol of Br₂ react with - 2/3 x 4 = 2.67 mol of Al

    but 3 mol of Al is present, more than the required amount.

    Therefore Al is the excess reactant

    Q3)

    according to the balanced equation mentioned previously,

    stoichiometry of Br₂ to AlBr₃ is 3:2

    the amount of product formed depends on amount of limiting reactant present.

    when 3 mol of Br₂ reacts - then 2 mol of AlBr₃ is formed

    now since 4 mol of the limiting reactant is present

    then when 4 mol of Br₂ reacts - 2/3x 4 = 2.67 mol of AlBr₃ is formed

    number of AlBr₃ moles formed are - 2.67 mol

    Q4)

    limiting reactant the whole amount is consumed.

    when Br₂ is the limiting reactant, 4 moles of Br₂ present are fully used up.

    as calculated above, 4 mol of Br₂ reacts with 2.67 mol of Al

    and 3 moles of Al provided.

    Used amount of Al - 2.67 mol

    Therefore amount of Al in excess = 3 - 2.67 = 0.33 mol

    amount of excess reactant left over after the reaction is 0.33 mol
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