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7 November, 06:52

When ethane (C2H6) burns, it produces carbon dioxide and water:

2C2H6 (g) + 7O2 (g) →4CO2 (g) + 6H2O (l)

How many liters of carbon dioxide will be produced when 89.5 L of ethane are burned?

(One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)

179 L

313 L

358 L

538 L

+5
Answers (1)
  1. 7 November, 06:57
    0
    179 L

    Explanation:

    We are given;

    The equation for the combustion of ethane.;

    2C₂H₆ (g) + 7O₂g) →4CO₂ (g) + 6H₂O (l)

    Volume of ethane gas used as 89.5 L

    We are required to determine the volume of carbon dioxide produced;

    Step 1: Determine the number of moles of ethane gas

    According to the molar gas volume, 1 mole of a gas occupies a volume of 22.4 L at STP.

    Therefore;

    Number of moles of ethane = Volume given/molar gas volume

    Thus;

    Moles of ethane = 89.5 L : 22.4 L

    = 3.9955

    = 4.0 moles

    Step 2: Determine the number of moles of CO₂ produced

    From the equation,

    12 moles of ethane reacts to produce 4 moles of carbon dioxide

    Thus; Moles of CO₂ = Moles of ethane * 2

    = 4.0 moles * 2

    = 8.0 moles

    Step 3: Volume of CO₂ produced

    We know that;

    1 mole of a gas = 22.4 L at STP

    Therefore;

    Volume of CO₂ = 8.0 moles * 22.4 L/mol

    = 179.2 L

    = 179 L

    Therefore, the volume of CO₂ is 179 L
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