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3 December, 20:02

Sodium metal and water react to form hydrogen and sodium hydroxide. If 11.96 g of sodium react with water to form 0.52 g of hydrogen and 20.80 g of sodium hydroxide, what mass of water was involved in the reaction

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  1. 3 December, 20:23
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    9.36g of H2O.

    Explanation:

    We'll begin by writing the balanced equation for the reaction.

    2Na + 2H2O - > 2NaOH + H2

    Next, we shall determine the mass of Na and H2O that reacted from the balanced equation.

    This is illustrated below:

    Molar mass of Na = 23g/mol

    Mass of Na from the balanced equation = 2 x 23 = 46g

    Molar mass of H2O = (2x1) + 16 = 18g/mol

    Mass of H2O from the balanced equation = 2 x 18 = 36g

    From the balanced equation above,

    46g of Na reacted with 36g of H2O.

    Now, we can determine the mass of H2O involved in reaction as follow:

    From the balanced equation above,

    46g of Na reacted with 36g of H2O.

    Therefore, 11.96g of Na will react with = (11.96 x 36) / 46 = 9.36g of H2O.

    Therefore, 9.36g of H2O were used for the reaction.
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