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12 February, 11:57

Consider the following unbalanced reaction for the combustion of hexane: C6H14 (g) + O2 (g) - -> CO2 (g) + H2O (g)

Balance the equation and determine how many grams of CO 2 gas will form when 5.35 grams of C6H14 reacts with excess O2.

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Answers (2)
  1. 12 February, 15:00
    0
    1. 2C6H14 + 19O2 - > 12CO2 + 14H2O

    2. 16.42g

    Explanation:

    1. We'll begin by balancing the equation. This is illustrated below:

    C6H14 + O2 - > CO2 + H2O

    There are 6 atoms of C on the left side and 1 atom on the right. It can be balance by putting 6 in front of CO2 as shown below:

    C6H14 + O2 - > 6CO2 + H2O

    There are 14 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 7 in front of H2O as shown below:

    C6H14 + O2 - > 6CO2 + 7H2O

    Now, there are a total of 19 atoms of O on the right side and 2 at on the left side. It can be balance by putting 19/2 in front of O2 as shown below:

    C6H14 + 19/2O2 - > 6CO2 + 7H2O

    Now multiply through by 2 to clear the fraction as shown below:

    2C6H14 + 19O2 - > 12CO2 + 14H2O

    Now the equation is balanced

    2. 2C6H14 + 19O2 - > 12CO2 + 14H2O

    Let us determine the mass of C6H14 that reacted and the mass of CO2 produced from the balanced equation.

    This is illustrated below:

    Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

    Mass of C6H14 from the balanced equation = 2 x 86 = 172g

    Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

    Mass of CO2 from the balanced equation = 12 x 44 = 528g

    From the equation above,

    172g of C6H14 produced 528g of CO2.

    Therefore, 5.35g of C6H14 will produce = (5.35 x 528) / 172 = 16.42g of CO2.

    From the calculations made above, 16.42g of CO2 will be produced from 5.35g of C6H14
  2. 12 February, 15:37
    0
    16.4 grams of Carbon dioxide (CO2) gas will be formed

    2C6H14 (g) + 19O2 (g) - -> 12CO2 (g) + 14H2O (g)

    Explanation:

    Step 1: Data given

    Mass of C6H14 = 5.35 grams

    Molar mass C6H14 = 86.18 g/mol

    Molar mass CO2 = 44.01 g/mol

    Step 2: The unbalanced equation

    C6H14 (g) + O2 (g) - -> CO2 (g) + H2O (g)

    Step 3: The balanced equation

    2C6H14 (g) + 19O2 (g) - -> 12CO2 (g) + 14H2O (g)

    Step 4: Calculate moles hexane (C6H14)

    Moles hexane = mass hexane / molar mass hexane

    Moles hexane = 5.35 grams / 86.18 g/mol

    Moles hexane = 0.0621 moles

    Step 5: Calculate moles CO2

    For 2 moles hexane we need 19 moles oxygen to produce 12 moles carbon dioxide and 14 moles water

    For 0.0621 moles hexane we'll have 6*0.0621 = 0.3726 moles CO2

    Step 6: Calculate mass CO2

    Mass CO2 = 0.3726 moles * 44.01 g/mol

    Mass CO2 = 16.4 grams

    16.4 grams of Carbon dioxide (CO2) gas will be formed
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