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9 May, 00:18

What is the mass of heavy water, D2O (l), produced when 7.60 g of O2 (g) reacts with excess D2 (g) ?

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  1. 9 May, 03:01
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    19.026 g

    Explanation:

    The reaction between D₂ and O₂ is given by the equation;

    2D₂ (g) + O₂ (g) → 2H₂O (l)

    We are given;

    Mass of oxygen (O₂) that reacted as 7.60 g

    We are required to calculate the mass of D₂O produced;

    Step 1: Calculate the moles of O₂ used (limiting reactant)

    To calculate the number of moles we divide mass by the molar mass.

    Moles = mass : Molar mass

    Molar mass of O₂ is 16.0 g/mol

    Therefore;

    Moles = 7.60 g : 16.0 g/mol

    = 0.475 moles

    Step 2: Calculate the moles of D₂O produced

    From the equation, 1 mole of oxygen reacts to produce 2 moles of D₂O

    Thus, the mole ratio of O₂ : D₂O is 1 : 2

    Therefore, moles of D₂O = Moles of O₂ * 2

    = 0.475 moles * 2

    = 0.95 moles

    Step 3: Calculate the mass of heavy water produced

    Mass = Number of moles * Molar mass

    Molar mass of heavy water = 20.0276 g/mol

    Therefore;

    Mass of heavy water = 0.95 moles * 20.0276 g/mol

    = 19.026 g

    Hence, the mass of heavy water produced is 19.026 g
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