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27 November, 04:30

Propane (C3H8) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. Include all reaction states. → (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.65 g of propane. L

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  1. 27 November, 07:24
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    C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

    11.6 L of CO2 will be produced

    Explanation:

    Step 1: Data given

    Mass of propane = 7.65 grams

    Molar mass propane = 44.1 g/mol

    Burning = combustion reation = adding O2. The products will be CO2 and H2O

    Step 2: The balanced equation

    C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

    Step 3: Calculate moles propane

    Moles propane = 7.65 grams / 44.1 g/mol

    Moles propane = 0.173 moles

    Step 4: Calculate moles CO2

    For 1 mol propane we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

    For 0.173 moles propane we'll have 3*0.173 = 0.519 moles CO2

    Step 5: Calculate volume of CO2

    1 mol = 22.4 L

    0.519 moles = 22.4 L * 0.519 = 11.6 L

    11.6 L of CO2 will be produced
  2. 27 November, 07:44
    0
    11.6 L will be the number of liters of carbon dioxide measured at STP.

    Explanation:

    The balanced equation for this combustion reaction is:

    C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

    where 1 mol of propane reacts to 5 moles of oxygen in order to produce 3 moles of carbon dioxide and 4 moles of water.

    We assume the oxygen in excess, so the limiting reagent is the propane. Now, we determine the moles: 7.65 g. 1 mol / 44 g = 0.174 moles

    Ratio is 1:3. 1 mol of propane can produce 3 moles of CO₂

    Therefore, 0.174 moles will produce (0.174. 3) / 1 = 0.521 moles of CO₂

    As 1 mol of gas is contained in 22.4L at STP conditions, we propose

    22.4L / 1 mol = V₂ / 0.521 mol

    22.4 L / 1 mol. 0.521 mol = V₂ → 11.6 L
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