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9 November, 14:37

Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al (s) + 2 NaOH (aq) + 6 H2O (l) → 2 NaAl (OH) 4 (aq) + 3 H2 (g) What t mass of hydrogen gas would be formed from a reaction of 1.83 g Al and 4.30 g NaOH in water?

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  1. 9 November, 15:50
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    The mass of hydrogen gas formed is 0.205 grams

    Explanation:

    Step 1: Data given

    Mass of 1.83 grams of Al

    Mass of NaOH = 4.30 grams

    Molar mass of Al = 26.98 g/mol

    Molar mass of NaOH = 40 g/mol

    Step 2: The balanced equation:

    2 Al (s) + 2 NaOH (aq) + 6 H2O (l) → 2 NaAl (OH) 4 (aq) + 3 H2 (g)

    Step 3: Calculate moles of Al

    Moles Al = mass Al / Molar mass Al

    Moles Al = 1.83 grams / 26.98 g/mol

    Moles Al = 0.0678 moles

    Step 4: Calculate moles of NaOH

    Moles NaOH = 4.30 grams / 40 g/mol

    Moles NaOH = 0.1075 moles

    Step 5: Calculate limiting reactant

    For 2 moles of Al, we need 2 moles of NaOH

    Aluminium is the limiting reactant. It will completely be consumed (0.0678 moles)

    NaOH is in excess. There will react 0.0678 moles

    There will remain 0.1075 - 0.0678 = 0.0397 moles

    Step 6: Calculate moles of hydrogen

    For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

    For 0.0678 moles of Al, there is produced 0.0678 * 3/2 = 0.1017 moles of H2

    Step 7: Calculate mass of H2

    Mass of H2 = Moles H2 * Molar mass of H2

    Mass of H2 = 0.1017 moles * 2.02 g/mol

    Mass of H2 = 0.205 grams

    The mass of hydrogen gas formed is 0.205 grams
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