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20 March, 14:09

Write and balance the combustion equation for propane. (Propane is combusted in the presence of oxygen to produce carbon dioxide and water). 2) How many grams of oxygen are required to burn 200 grams of propane

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  1. 20 March, 14:17
    0
    363.64g of oxygen would be required.

    Explanation:

    1) Write and balance the combustion equation for propane.

    Propane + Oxygen - -> carbon dioxide + water

    C3H8 + O2 - -> CO2 + H2O

    Upon balancing, we have;

    C3H8 + 5O2 - -> 3CO2 + 4H2O

    2) How many grams of oxygen are required to burn 200 grams of propane.

    From the reaction;

    Propane = (3 * 12) + (8 * 1) = 44

    Oxygen = (5 * 16) = 80

    80 grams of oxygen is required to combust 44g of propane.

    80 = 44

    x = 200

    x = (80 * 200) / 44

    x = 363.64g
  2. 20 March, 14:36
    0
    C3H8 + 5O2 → 3CO2 + 4H2O

    We need 725.6 grams O2 to burn 200 grams of propane

    Explanation:

    Step 1: Data given

    propane = C3H8

    Molar mass propane = 44.1 g/mol

    Mass of propane = 200.0 grams

    Molar mass of O2 = 32.0 g/mol

    Step 2: The balanced equation

    C3H8 + 5O2 → 3CO2 + 4H2O

    Step 3: Calculate moles propane

    Moles propane = mass propane / molar mass propane

    Moles propane = 200.0 grams / 44.1 g/mol

    Moles propane = 4.535 moles

    Step 4: Calculate moles O2

    For 1 mol C3H8 we need 5 moles O2 to react to produce 3 moles CO2 and 4 moles H2O

    For 4.535 moles propane we need 5*4.535 = 22.675 moles O2

    Step 5: Calculate mass O2

    Mass O2 = moles O2 * molar mass O2

    Mass O2 = 22.675 moles * 32.0 g/mol

    Mass O2 = 725.6 grams

    We need 725.6 grams O2 to burn 200 grams of propane
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