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27 May, 19:39

A student in chemistry 150-02 weighed out 55.5 g of octane (C8H18) and allowed it to react with oxygen, O2. The products formed were carbon dioxide (CO2) and water (H2O).

Write a balanced equation for the reaction

How many grams of oxygen are required to react with 55.0g of octane (C8H18) ?

How many grams of CO2 are produced from 55.0g of octane (C8H18) ?

How many molecules of H2O are produced from 55.0g of octane (C8H18) ?

How many gams of C8H18 are required to produce 30.0g of water (H2O).?

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Answers (1)
  1. 27 May, 21:49
    0
    Given dа ta:

    Mass of octane = 55.5 g

    Balanced chemical equation = ?

    Mass of oxygen required to react = ?

    Mass of CO₂ for med = ?

    Molecules of water produced = ?

    Mass of octane required to produced 30.0 g of water = ?

    Solution:

    1)

    Chemical equation:

    2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

    2)

    Mass of oxygen required to react = ?

    Mass of octane = 55.0 g

    Solution:

    Number of moles of octane:

    Number of moles = mass / molar mass

    Number of moles = 55.0 g/114.23 g/mol

    Number of moles = 0.48 mol

    Now we will compare the moles of octane with oxygen.

    C₈H₁₈ : O₂

    2 : 25

    0.48 : 25/2*0.48 = 6 mol

    Mass of oxygen required:

    Mass = number of moles * molar mass

    Mass = 6 mol * 32 g/mol

    Mass = 192 g

    3)

    Given dа ta:

    Mass of carbon dioxide produced = ?

    Mass of octane = 55g

    Solution:

    Number of moles of octane:

    Number of moles = mass / molar mass

    Number of moles = 55.0 g/114.23 g/mol

    Number of moles = 0.48 mol

    Now we will compare the moles of octane with CO₂.

    C₈H₁₈ : CO₂

    2 : 16

    0.48 : 16/2*0.48 = 3.84 mol

    Mass of CO₂ produced:

    Mass = number of moles * molar mass

    Mass = 3.84 mol * 44 g/mol

    Mass = 168.96 g

    4)

    Given dа ta:

    Molecules of water produced = ?

    Mass of octane = 55g

    Solution:

    Number of moles of octane:

    Number of moles = mass / molar mass

    Number of moles = 55.0 g/114.23 g/mol

    Number of moles = 0.48 mol

    Now we will compare the moles of octane with H₂O.

    C₈H₁₈ : H₂O

    2 : 18

    0.48 : 18/2*0.48 = 4.32 mol

    Number of molecules of water:

    1 mol = 6.022 * 10²³ molecules

    4.32 mol * 6.022 * 10²³ molecules / 1 mol

    26 * 10²³ molecules

    5)

    Given dа ta:

    Mass of octane required = ?

    Mass of water produced = 30 g

    Solution:

    Number of moles of water.

    Number of moles = mass / molar mass

    Number of moles = 30 g / 18 gmol

    Number of moles = 1.67 mol

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

    2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

    H₂O : C₈H₁₈

    18 : 2

    1.67 : 2/18*1.67 = 0.185 mol

    Mass of octane:

    Mass = number of moles * molar mass

    Mass = 0.185 * 114.23 g/mol

    Mass = 21.13 g
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