Chapter 9, Problem 49A

Interpretation Introduction

Interpretation:

The balanced combustion reaction of glucose needs to be determined and the theoretical yield of carbon dioxide when 1.00 g of glucose is burned completely needs to be determined.

Concept Introduction:

When an atom combines with oxygen and it produces carbon dioxide and water, thus, the reaction is known as combustion reaction

The amount or product which is theoretically calculated from the amount of limiting agent is known as theoretical yield.

Answer to Problem 49A

The combustion reaction is shown below:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{+}{\text{6O}}_{\text{2}}\left(\text{g}\right)\stackrel{\text{Δ}}{\to }{\text{6CO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}$

The theoretical yield is 1.32 g

Explanation of Solution

The balance combustion reaction of glucose is shown below:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{+}{\text{6O}}_{\text{2}}\left(\text{g}\right)\stackrel{\text{Δ}}{\to }{\text{6CO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}$

When glucose reacts with oxygen in presence of heat then it will produce carbon dioxide and water which is known as combustion reaction.

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{=}\frac{\text{mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{\text{molar}\text{mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\\ \text{=}\frac{\text{1}\text{.00}\text{g}}{\text{180}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.005}\text{mol}\end{array}$

1 mol of C₆H₁₂O₆ = 6 mol of CO₂

0.005 mol of C₆H₁₂O₆ = 6 × 0.005 mol of CO₂

= 0.03 mol of CO₂

The calculation of theoretical yield is shown below:

Theoretical yield = moles × molar mass

= 0.03 × 44 gmol-1

= 1.32 g

Conclusion

The combustion reaction is shown below:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{+}{\text{6O}}_{\text{2}}\left(\text{g}\right)\stackrel{\text{Δ}}{\to }{\text{6CO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}$

The theoretical yield is 1.32 g