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22 March, 22:48

What is the total number of moles of aluminum oxide that can be formed when 54 grams of aluminum reacts completely with oxygen?

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  1. 23 March, 00:10
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    4Al + 3O₂ → 2Al₂O₃

    m (Al) = 54 g

    M (Al) = 27 g/mol

    n (Al₂O₃) = n (Al) / 2

    n (Al) = m (Al) / M (Al)

    n (Al₂O₃) = m (Al) / {2M (Al) }

    n (Al₂O₃) = 54/{2*27} = 1 mol
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