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16 April, 02:15

For the following reaction, 7.43 grams of methane (CH4) are allowed to react with 27.6 grams of carbon tetrachloride. methane (CH4) (g) + carbon tetrachloride (g) dichloromethane (CH2Cl2) (g) What is the maximum amount of dichloromethane (CH2Cl2) that can be formed

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  1. 16 April, 02:32
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    30.4 grams of CH2Cl2 will be formed

    Explanation:

    Step 1: data given

    Mass of methane = 7.43 grams

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

    Mass of CCl4 = 27.6 grams

    Molar mass of CCl4 = 153.82 g/mol

    Step 2: The balanced equation

    CH4 + CCl4 → 2CH2Cl2

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles methane = mass methane / molar mass methane

    Moles methane = 7.43 grams / 16.04 g/mol

    Moles methane = 0.463 moles

    Moles CCl4 = 27.6 grams / 153.82 g/mol

    Moles CCl4 = 0.179 moles

    Step 4: Calculate limiting reactant

    For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

    CCl4 is the limiting reactant. It will completely be consumed. (0.179 moles). Ch4 is in excess. There will react 0.179 moles. There will remain 0.463 - 0.179 = 0.284 moles

    Step 5: Calculate moles CH2Cl2

    For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

    For 0.179 moles CCl4 we'll have 2*0.179 = 0.358 moles CH2Cl2

    Step 6: Calculate mass CH2Cl2

    Mass CH2Cl2 = 0.358 moles * 84.93 g/mol

    Mass CH2Cl2 = 30.4 grams

    30.4 grams of CH2Cl2 will be formed
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