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6 October, 19:31

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.28 g of octane is mixed with 15. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

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  1. 6 October, 22:15
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    Mass of water will be 3.24 g.

    Explanation:

    Given dа ta:

    mass of octane = 2.28 g

    mass of oxygen = 15 g

    mass of water = ?

    Solution:

    First of all we will write the balance chemical equation.

    2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

    Now we will calculate the moles of octane and oxygen.

    number of moles of octane = mass / molar mass

    number of moles of octane = 2.28 g / 114.33 g/mol

    number of moles of octane = 0.02 mol

    number of moles of oxygen = mass / molar mass

    number of moles of oxygen = 15 g / 32 g/mol

    number of moles of oxygen = 0.47 mol

    Now we will compare the moles of oxygen and octane with water from balance chemical equation

    C₈H₁₈ : H₂O

    2 : 18

    0.02 : 18/2*0.02 = 0.18 mol

    O₂ : H₂O

    25 : 18

    0.47 : 18/25*0.47 = 0.34 mol

    Number of moles of water produced by octane is less that's why octane will be the limiting reactant.

    Mass of water = number of moles * molar mass

    Mass of water = 0.18 mol * 18 g/mol = 3.24 g
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