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11 February, 13:35

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) If 0.391 g of carbon dioxide is produced from the reaction of 0.16 g of methane and 0.84 g of oxygen gas, calculate the percent yield of carbon dioxide.

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  1. 11 February, 13:39
    0
    The % yied of carbon dioxide = 89.1 %

    Explanation:

    Step 1: Data given

    Mass of CO2 = 0.391 grams

    Mass of methane = 0.16 grams

    Mass of oxygen gas = 0.84 grams

    Molar mass of CO2 = 44.01 g/mol

    Molar mass of methane (CH4) = 16.04 g/mol

    Molar mass of oxygen gas (O2) = 32 g/mol

    Molar mass of H2O = 18.02 g/mol

    Step 2: The balanced equation

    CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

    Step 3: Calculate moles of CH4

    Moles CH4 = mass CH4 / Molar mass CH4

    Moles CH4 = 0.16 grams / 16.04 g/mol

    Moles CH4 = 0.00998 mol

    Step 4: Calculate moles of O2

    Moles O2 = mass O2 / Molar mass O2

    Moles O2 = 0.84 grams / 32 g/mol

    Moles O2 = 0.02625 mol

    Step 5: Calculate limiting reactant

    For 1 mole CH4 consumed, we need 2 moles O2, to produce 1 mole of CO2

    CH4 is the limiting reactant. It will completely be consumed (0.00998 moles)

    O2 is in excess. There will be consumed 2*0.00998 = 0.01996 mol

    There will remain 0.02625 - 0.01996 = 0.00629 mol O2

    Step 6: Calculate moles carbon dioxide

    For 1 mole CH4 consumed, we need 2 moles O2, to produce 1 mole of CO2

    For 0.00998 moles of CH4, we have 0.00998 moles of CO2

    Step 7: Calculate theoretical mass of CO2

    Mass of CO2 = moles of CO2 * Molar mass of CO2

    Mass of CO2 = 0.00998 moles * 44.01 g/mol

    Mass of CO2 = 0.439 grams

    Step 8: Calculate the % tield of CO2

    % yield = actual yield / theoretical yield

    % yield = (0.391 grams / 0.439 grams) * 100%

    % yield = 89.1%

    The % yied of carbon dioxide = 89.1 %
  2. 11 February, 13:40
    0
    percent yield of carbon dioxide is 46.5 %

    Explanation:

    Formula for the above reaction is

    CH₄+2O₂→CO₂+2H₂OCH₄+2O₂→CO₂+2H₂O

    Firstly I will calculate the moles of reactants

    Moles of CH₄ = given mass / Molar mass = 0.16g / 16gmole=0.01 moles of CH₄ (limiting reactant)

    Moles of Oxygen O₂ = given mass / Molar mass = 0.84 g/32gmole=0.026 moles of O₂ (reactant in excess)

    Thus, moles of CO₂ = 0.01

    Mass of CO₂ = 0.01 mole * 44 g mole=0. 44 grams

    Therefore,

    For percent yield=experimental yield / theoretical yield*100%.

    % yield = 0.391g/0.84*100%=46.5 %
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