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1 December, 08:43

Suppose 6.87 g of sulfuric acid is mixed with 9.7 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

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  1. 1 December, 12:25
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    9.96g of Na₂SO₄ is the maximum mass that could be produced

    Explanation:

    We must determine the reaction:

    Reactants: H₂SO₄, NaOH

    Products: H₂O, Na₂SO₄

    The equation is: H₂SO₄ (aq) + 2NaOH (aq) → 2H₂O (l) + Na₂SO₄ (aq)

    We have the mass of the reactants. We need to convert them to moles, in order to define the limiting reactant

    6.87 g / 98 g/mol = 0.0701 moles of acid

    9.7 g / 40 g/mol = 0.242 moles of base

    Limiting reactant is the acid. Let's verify

    2 moles of NaOH can react with 1 mol of acid

    Therefore 0.242 moles of NaOH must react with (0.242. 1) / 2 = 0.121 moles

    We do not have enough acid.

    Ratio with the salt is 1:1. 1 mol of acid produces 1 mol of salt

    Therefore 0.0701 moles of acid will produce 0.0701 moles of salt

    We convert the moles to mass → 0.0701 mol. 142.06 g / 1 mol = 9.96g
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