Chapter 11, Problem 64A

Interpretation Introduction

Interpretation:

The mass of octane needed to release 5.00 moles of carbon dioxide needs to be determined with the balanced chemical equation of combustion of octane.

Concept introduction:

The combustion reaction is an exothermic reaction which is the reaction of molecule with oxygen to form carbon dioxide and water. The energy released during the combustion is called as enthalpy of combustion.

Answer to Problem 64A

$\text{ 71}\text{.2 g octane }$

Explanation of Solution

The combustion of octant occurs in the presence of oxygen and leads to the formation of carbon dioxide and water. Hence the reaction must be :

${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{(g)+O}}_{\text{2(g)}}\stackrel{}{\to }{\text{CO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(g)}$

Now we have to balance it by balancing all atoms at both sides. Start with balancing the number of C atoms at both side:

${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{(g)+O}}_{\text{2(g)}}\stackrel{}{\to }{\text{16CO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(g)}$

Now balance O atoms:

${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{(g)+25O}}_{\text{2(g)}}\stackrel{}{\to }{\text{16CO}}_{\text{2}}{\text{(g)+18H}}_{\text{2}}\text{O(g)}$

Since the H atoms are balance now hence the balance chemical equation for the combustion of octane is:

${\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}{\text{(g)+25O}}_{\text{2(g)}}\stackrel{}{\to }{\text{16CO}}_{\text{2}}{\text{(g)+18H}}_{\text{2}}\text{O(g)}$

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation.

Moles of carbon dioxide = 5.00 moles

16 moles of carbon dioxide = 2 mole of octane

Molar mass of octane = 114 g/mol

Calculate mass of octane with the help of mole ratio:

$\text{5}{\text{.00 moles CO}}_{\text{2}}\text{ ×}\frac{\text{2 moles octane }}{{\text{16 moles CO}}_{\text{2}}}\text{×}\frac{\text{114 g octane }}{\text{1 mole octanen }}\text{= 71}\text{.2 g octane }$

Conclusion

Mole ratio is used to calculate the mass of product.