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25 December, 00:02

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide, calculate the percent yield of sodium sulfate

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  1. 25 December, 00:07
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    The % yield is 27.0 %

    Explanation:

    Step 1: Data given

    Mass of sulfuric acid = 4.9 grams

    Mass of sodium hydroxide = 7.8 grams

    Mass of sodium sulfate produced = 1.92 grams

    Molar mass H2SO4 = 98.08 g/mol

    Molar mass NaOH = 40 g/mol

    Molar mass Na2SO4 = 142.04 g/mol

    Step 2: The balanced equation

    H2SO4 + 2NaOH → Na2SO4 + 2H2O

    Step 3: Calculate moles H2SO4

    Moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

    Moles H2SO4 = 4.9 grams / 98.08 g/mol =

    Moles H2SO4 = 0.05 moles

    Step 4: Calculate moles NaOH

    Moles NaOH = 7.8 grams / 40 g/mol

    Moles NaOH = 0.195 moles

    Step 5: Calculate limiting reactant

    For 1 mole H2SO4 consumed, we need 2 moles NaOH to produce 1 mole Na2SO4 and 2 moles H2O

    H2SO4 is the limiting reactant. It will completely be consumed (0.05 moles).

    NaOH is in excess. There will react 2*0.05 = 0.1 moles

    There will remain 0.195 - 0.1 = 0.095 moles NaOH

    Step 6: Calculate moles Na2SO4

    For 1 mole H2SO4 consumed, we need 2 moles NaOH to produce 1 mole Na2SO4

    For 0.05 moles H2SO4, we have 0.05 moles Na2SO4

    Step 7: Calculate mass of Na2SO4

    Mass Na2SO4 = Moles Na2SO4 * Molar mass Na2SO4

    Mass = 0.05 moles * 142.04 g/mol = 7.102

    This is the theoretical yield

    Step 8: Calculate the percent yield of Na2SO4

    % yield = (actual yield / theoretical yield) * 100%

    % yield = (1.92 / 7.102) * 100% = 27.0 %

    The % yield is 27.0 %
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