Question

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.

Answer 1

Answer:

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

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

Step-by-step 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

Answer 2

1. 2C6H14 + 19O2 —> 12CO2 + 14H2O

2. 16.42g

Step-by-step 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