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27 May, 06:37

Aqueous sulfuric acid (H2SO4) reacts with solid sodium hydroxide (NaOH) to produce aqeous sodium sulfate (Na2SO4) and liquid water (H2O). what is the theoretical yield of water formed from the reaction of 5.9 g of sulfuric acid and 6.6 g of sodium hydroxide? Be sure your answer has the correct number of significant digits in it.

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  1. 27 May, 10:22
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    The theoretical yield of water formed is 2.2 grams

    Explanation:

    Step 1: Data given

    Mass of H2SO4 = 5.9 grams

    Mass of NaOH = 6.6 grams

    Molar mass H2SO4 = 98.08 g/mol

    Molar mass of NaOH = 40.0 g/mol

    Step 2: The balanced equation

    2NaOH + H2SO4 → Na2SO4 + 2H2O

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles H2SO4 = 5.9 grams / 98.08 g/mol

    Moles H2SO4 = 0.060 moles

    Moles NaOH = 6.6 grams / 40.0 g/mol

    Moles NaOH = 0.165 moles

    Step 4: Calculate the limiting reactant

    For 2 moles NaOH we need 1 mol H2SO4 to produce 1 mol Na2SO4 and 2 moles H2O

    H2SO4 is the limiting reactant. It will completely be consumed (0.060 moles). NaOH is in excess. There will react 2*0.060 = 0.120 moles

    There will remain 0.165 - 0.120 = 0.045 moles NaOH

    Step 5: Calculate moles H2O

    For 2 moles NaOH we need 1 mol H2SO4 to produce 1 mol Na2SO4 and 2 moles H2O

    For 0.0600 moles H2SO4 we'll have 2*0.0600 = 0.120 moles H2O

    Step 6: Calculate mass H2O

    Mass H2O = 0.120 moles * 18.02 g/mol

    Mass H2O = 2.16 grams

    The theoretical yield of water formed is 2.2 grams
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