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5 August, 05:00

If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according to the following chemical equation? C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)

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Answers (2)
  1. 5 August, 05:25
    0
    9.94 mL, the volume of ethanol needed

    Explanation:

    The reaction is:

    C₂H₅OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 3 H₂O (l)

    We convert the mass of the formed product to moles:

    15 g. 1mol / 44g = 0.341 moles

    2 moles of dioxide are produced by 1 mol of ethanol, in order to stoichiometry.

    Therefore, 0.341 moles of CO₂ must be produced by (0.341. 1) / 2 = 0.1705 moles of alcohol.

    We convert the moles to mass, and then, the mass to volume by the use of density.

    0.1705 mol. 46 g / 1 mol = 7.84 g of ethanol

    Ethanol density = Ethanol mass / Ethanol volume

    Ethanol volume = Ethanol mass / Ethanol density → 7.84 g / 0.789 g/mL =

    9.94 mL
  2. 5 August, 05:29
    0
    We need 9.95 mL of ethanol

    Explanation:

    Step 1: Data given

    Density ethanol = 0.789 g/mL

    Mass CO2 produced = 15.0 grams

    Step 2: The balanced equation

    C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)

    Step 3: Calculate moles CO2

    Moles CO2 = mass CO2 / molar mass CO2

    Moles CO2 = 15.0 grams / 44.01 g/mol

    Moles CO2 = 0.341 moles

    Step 4: Calculate moles ethanol

    For 1 mol ethanol, 3 moles we need O2 to produce 2 moles CO2 and 3 moles H2O

    For 0.341 moles CO2 we need 0.341 / 2 = 0.1705 moles ethanol

    Step 5: Calculate mass ethanol

    Mass ethanol = moles ethanol * molar mass ethanol

    Mass ethanol = 0.1705 moles * 46.07 g/mol

    Mass ethanol = 7.85 grams

    Step 6: Calculate volume ethanol

    Volume ethanol = mass / density

    Volume = 7.85 grams / 0.789 g/mL

    Volume = 9.95 mL

    We need 9.95 mL of ethanol
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