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24 December, 21:13

Gaseous methane CH4 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 12.4g of carbon dioxide is produced from the reaction of 7.38g of methane and 39.2g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures.

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  1. 24 December, 21:38
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    The percent yield = 61.8 %

    Explanation:

    Step 1: Data given

    Mass of CO2 produced = 12.4 grams

    Mass of methane = 7.38 grams

    Mass of oxygen = 39.2 grams

    Molar mass of CO2 = 44.01 g/mol

    Molar mass of methane = 16.04 g/mol

    Molar mass of oxygen = 32.0 g/mol

    Step 2: The balanced equation

    CH4 + 2O2 → CO2 + 2H2O

    Step 3: Calculate moles methane

    Moles methane = mass methane / molar mass methane

    Moles methane = 7.38 grams / 16.04 g/mol

    Moles methane = 0.460 moles

    Step 4: Calculate moles oxygen

    Moles O2 = 39.2 grams / 32.0 g/mol

    Moles O2 = 1.225 moles

    Step 5: Calculate limiting reactant

    For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

    Methane is the limiting reactant. It will completely be consumed (0.460 moles). O2 is in excess. There will react 2*0.460 = 0.920 moles

    There will remain 1.225 - 0.920 = 0.305 moles O2

    Step 6: Calculate moles of CO2

    For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

    For 0.460 moles CH4 we'll have 0.460 moles CO2

    Step 7: Calculate mass CO2

    Mass CO2 = moles CO2 * molar mass CO2

    Mass CO2 = 0.460 moles * 44.01 g/mol

    Mass CO2 = 20.24 grams

    Step 8: Calculate percent yield

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

    % yield = (12.5 grams / 20.24 grams) * 100%

    % yield = 61.8 %

    The percent yield = 61.8 %
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