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4 October, 11:22

The combustion of methyl alcohol in an abundant excess of oxygen follows the equation:

2CH3OH + 3O2? 2CO2 + 4H2O

When 3.85 g of CH3OH was mixed with 6.15 g of O2 and ignited, 3.65 g of CO2 was obtained. What was the percentage yield of CO2?

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  1. 4 October, 12:32
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    The % yield of CO2 is 69.1 %

    Explanation:

    Step 1: Data given

    Mass of CH3OH = 3.85 grams

    Mass of O2 = 6.15 grams

    Mass of CO2 obtained = 3.65 grams

    Molar mass of CH3OH = 32.04 g/mol

    Molar mass O2 = 32.00 g/mol

    Molar mass CO2 = 44.01 g/mol

    Step 2: The balanced equation

    2CH3OH + 3O2→ 2CO2 + 4H2O

    Step 3: Calculate moles CH3OH

    Moles CH3OH = mass CH3OH / molar mass CH3OH

    Moles CH3OH = 3.85 grams / 32.04 g/mol

    Moles CH3OH = 0.120 moles

    Step 4: Calculate moles O2

    Moles O2 = 6.15 grams / 32.00 g/mol

    Moles O2 = 0.192 moles

    Step 5: Calculate limtiting reactant

    CH3OH is the limiting reactant. It will completely be consumed (0.120 moles). O2 is in excess. There will react 3/2 * 0.120 = 0.180 moles

    There will remain 0.192 - 0.180 = 0.012 moles

    Step 6: Calculate moles CO2

    For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

    For 0.120 moles CH3OH we'll have 0.120 moles CO2

    Step 7: Calculate mass CO2

    Mass CO2 = moles CO2 * molar mass CO2

    Mass CO2 = 0.120 moles * 44.01 g/mol

    Mass CO2 = 5.28 grams

    Step 8: Calculate % yield

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

    %yield = (3.65 grams / 5.28 grams) * 100%

    % yield = 69.1 %

    The % yield of CO2 is 69.1 %
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