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17 January, 13:26

Al (s) + O2 (g) Al2O3 (s) (a) Consider the unbalanced equation above. How many grams of O2 are required to react with 49.0 g of aluminum? Use at least as many significant figures in your molar masses as in the data given.

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  1. 17 January, 16:27
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    We need 43.58 grams of O2 to react with 49.0 grams of aluminium

    Explanation:

    Step 1: Data given

    Mass of aluminium = 49.0 grams

    Molar mass of aluminium = 26.98 g/mol

    Step 2: The balanced equation

    4Al (s) + 3O2 (g) → 2Al2O3 (s)

    Step 3: Calculate moles aluminium

    Moles aluminium = mass aluminium / molar mass aluminium

    Moles aluminium = 49.0 grams / 26.98 g/mol

    Moles aluminium = 1.816 moles

    Step 4: Calculate moles O2 needed

    For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

    For 1.816 moles aluminium we need 3/4 * 1.816 = 1.362 moles O2

    Step 5: Calculate mass O2

    Mass O2 = moles O2 * molar mass O2

    Mass O2 = 1.362 moles * 32.0 g/mol

    Mass O2 = 43.58 grams

    We need 43.58 grams of O2 to react with 49.0 grams of aluminium
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