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1 November, 13:56

In a double replacement reaction copper (II) chloride combines with sodium nitrate. When a student combines 3 grams of copper (II) chloride react with 4 grams of sodium nitrate, how much sodium chloride is produced?

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  1. 1 November, 14:07
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    0.0446 moles of NaCl are produced by the reaction

    Explanation:

    First of all we convert the mass of reactants to moles, to determine the limiting reagent

    3 g / 134.45 g/mol = 0.0223 moles of CuCl₂

    4 g / 85 g/mol = 0.0470 moles of NaNO₃

    The reaction is: CuCl₂ (aq) + 2NaNO₃ (aq) → 2NaCl (aq) + Cu (NO₃) ₂ (aq)

    By stoichiomety, 1 mol of chloride reacts with 2 moles of nitrate

    Then, 0.0223 moles of chloride will react with (0.0223. 2) / 1 = 0.0446 moles of nitrate. We have 0.0470 moles, so the nitrate is in excess, and the limiting reactant is the chloride.

    2 moles of nitrate react with 1 mol of chloride

    Then, 0.0470 moles of nitrate will react with (0.0470. 1) / 2 = 0.0235 moles of chloride. There is not enough chloride because we have 0.0223 moles and we need 0.0235.

    In the reaction, 1 mol of chloride produces 2 mol of sodium chloride

    Then, our 0.0223 moles will produce (0.0223.2) / 1 = 0.0446 moles of NaCl
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