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6 June, 19:21

Determine the limiting reactant. 2al (s) + 3br2 (g) →2albr3 (s)

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  1. 6 June, 22:34
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    Missing question: For each of the following reactions, 25.0 g of each reactant is present initially.

    m (Al) = 25 g.

    n (Al) = m (Al) : M (Al).

    n (Al) = 25 g : 27 g/mol.

    n (Al) = 0,926 mol.

    m (Br₂) = 25 g.

    n (Br₂) = 25 g : 160 g/mol.

    n (Br₂) = 0,156 mol.

    Bromine is the limiting reactant, because it has less amount of substance.
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