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10 January, 13:17

For the following chemical reaction, determine the amount of water in grams produced when 2.05 x 10 molecules of H react with 1.33 x 10 molecules of O to produce water 2H (g) + O (g) → 2HO (l)

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  1. 10 January, 14:32
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    6.13 grams of H2O is produced

    Explanation:

    For the following chemical reaction, determine the amount of water in grams produced when 2.05 x 10^23 molecules of H2 react with 1.33 x 10^23 molecules of O2 to produce water

    Step 1: Data given

    Number of H2 molecules = 2.05 * 10^23 molecules

    Number of O2 molecules = 1.33 * 10^23 molecules

    Number of Avogadro = 6.022*10^23 / mol

    Molar mass H2 = 2.02 g/mol

    Molar mass O2 = 32.0 g/mol

    Molar mass H2O = 18.02 g/mol

    Step 2: The balanced equation

    2H2 (g) + O2 (g) → 2H2O (l)

    Step 3: Calculate moles

    Moles H2 = 2.05 * 10^23 / 6.022 * 10^23

    Moles H2 = 0.340 moles

    Step 4: Calculate moles O2

    Moles O2 = 1.33 * 10^23 molecules / 6.022*10^23

    Moles O2 = 0.221 moles

    Step 5: Calculate limiting reactant

    For 2 moles H2 we need 1 mol O2 to produce 2 moles H2O

    H2 is the limiting reactant. It will completely be consumed (0.340 moles).

    O2 is in excess. There will react 0.340/2 = 0.170 moles

    There will remain 0.221 - 0.170 = 0.051 moles

    Step 6: Calculate moles H2O

    For 2 moles H2 we need 1 mol O2 to produce 2 moles H2O

    For 0.340 moles we'll have 0.340 moles H2O

    Step 7: Calculate mass H2O

    Mass H2O = moles H2O * molar mass H2O

    Mass H2O = 0.340 moles * 18.02 g/mol

    Mass H2O = 6.13 grams

    6.13 grams of H2O is produced
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