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28 October, 10:08

Methanol (CH3OH) burns in air according to the equation 2CH3OH + 3O2 → 2CO2 + 4H2O If 201 g of methanol are used up in a combustion process, what is the mass of H2O produced?

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  1. 28 October, 10:29
    0
    221.6 g of H₂O are produced by this reaction

    Explanation:

    The combustion is this:

    2CH₃OH + 3O₂ → 2CO₂ + 4H₂O

    As we use 201 g of methanol, we assume the oxygen in excess.

    2 moles of methanol can produce 4 moles of water.

    We convert the grams of methanol to moles → 201 g. 1mol/32 g = 6.28 moles

    Let's make a rule of three:

    2 moles of methanol can produce 4 moles of water.

    6.28 moles of methanol must produce (6.28.4) / 2 = 12.5 moles of H₂O

    We convert the moles to mass → 12.5 mol. 18 g / 1mol = 226.1 g
  2. 28 October, 10:32
    0
    226.0 grams of H2O is produced

    Explanation:

    Step 1: Data given

    Mass of methanol = 201 grams

    Molar mass methanol = 32.04 g/mol

    Step 2: The balanced equation

    2CH3OH + 3O2 → 2CO2 + 4H2O

    Step 3: Calculate moles methanol

    Moles methanol = mass methanol / molar mass

    Moles methanol = 201.0 grams / 32.04 g/mol

    Moles methanol = 6.27 moles

    Step 4: Calculate moles H2O

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

    For 6.27 moles methanol we'll have 2*6.27 = 12.54 moles H2O

    Step 5: Calculate mass H2O

    Mass H2O = moles H2O * molar mass H2O

    Mass H2O = 12.54 moles * 18.02 g/mol

    Mass H2O = 226.0 grams

    226.0 grams of H2O is produced
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