Question

The equation for the combustion of glucose is: ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)+6{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$. How many grams of ${\mathrm{H}}_2\mathrm{O}$ will be produced when $8.064\mathrm{g}$ of glucose is burned?

Answer 1

Step 1: Combustion reaction :

• It refers to an exothermic chemical reaction that has a high temperature.
• It happens between a fuel and an oxidant.
• A combustion reaction in a fire produces a flame, and the heat emitting can thus make it self-sustaining.
• A combustion reaction is often a complicated sequence of elementary radical reactions.

Step 2: Given information:

• The amount of glucose is $8.064\mathrm{g}$.
• The molar mass of glucose is $180.15588\mathrm{g}{\mathrm{mol}}^{-1}$.
• The molar mass of water is $18.01528\mathrm{g}{\mathrm{mol}}^{-1}$.

Step 3: Balanced chemical equation:

The equation for the combustion of glucose is given as:

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)+6{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Step 4: Calculation of the amount of water:

The following conversion will take place;

$$\begin{gathered} \Rightarrow 8.064\mathrm{g}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\times \frac{1\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}{180.15588\mathrm{g}{\mathrm{mol}}^{-1}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}\times \frac{6\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{mol}\mathrm{of}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}\times \frac{18.01528\mathrm{g}{\mathrm{mol}}^{-1}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}} \\ \mathrm{On}\mathrm{cancelling},\mathrm{the}\mathrm{amount}\mathrm{of}\mathrm{water}\mathrm{is}: \\ =4.838\mathrm{g}\mathrm{of}{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Thus, approximately $0.8\mathrm{g}$ water will produce.