Ask Question
11 November, 05:05

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.

+1
Answers (2)
  1. 11 November, 06:29
    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. 11 November, 07:06
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Consider the following unbalanced reaction for the combustion of hexane: C6H14 (g) + O2 (g) - -> CO2 (g) + H2O (g) Balance the equation and ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers