Question

The equation for the combustion of glucose is: ${{C}_{6}}{{H}_{12}}{{O}_{6}}\left( s \right)+6{{O}_{2}}\left( g \right)\to 6C{{O}_{2}}\left( g \right)+6{{H}_{2}}O\left( g \right)$
How many grams of ${{H}_{2}}O$ will be produced when 8.064 g of glucose is burned?

Answer 1

As we know that the combustion reaction of ethanol is a type of redox reaction, during which electrons are transferred between two reactants participating in it. We can identify this transfer of electrons by observing the changes in the oxidation states of the reacting species.

Complete answer:
- Let us firstly discuss the combustion reaction. A combustion reaction is the reaction in which a substance responds with oxygen gas. And energy is delivered as light. In the combustion reaction oxygen is one of the reactants.
- When glucose is reacted with oxygen, then carbon dioxide gas is released along with water. We can say that this reaction is a redox reaction. The combustion reaction of glucose can be written as: ${{C}_{6}}{{H}_{12}}{{O}_{6}}\left( s \right)+6{{O}_{2}}\left( g \right)\to 6C{{O}_{2}}\left( g \right)+6{{H}_{2}}O\left( g \right)$
- As we know that the molar mass of glucose and water are 180.15588 g/mol and 18.01528 g/mol.
- We will get the value of molar mass of water by following steps:
Firstly, divide 8.064 g of glucose by its molar mass to get the moles of glucose, then will multiply times the mole ratio of $\frac{6\text{ }mol\text{ }{{H}_{2}}O}{1\text{ }mol\text{ }{{C}_{6}}{{H}_{12}}{{O}_{6}}}$to get the moles water. Then multiply the molar mass of water to get the mass of water. We can find the value as:
$\begin{aligned} & 8.064\text{ }g\text{ }{{C}_{6}}{{H}_{12}}{{O}_{6}}\times \frac{1\text{ }mol\text{ }{{C}_{6}}{{H}_{12}}{{O}_{6}}}{180.15588\text{ }g\text{ }{{C}_{6}}{{H}_{12}}{{O}_{6}}}\times \frac{6\text{ }mol\text{ }{{H}_{2}}O}{1\text{ }mol\text{ }{{C}_{6}}{{H}_{12}}{{O}_{6}}}\times \frac{18.01528\text{ }g\text{ }{{H}_{2}}O}{\text{1 }mol\text{ }{{H}_{2}}O} \\\ & =4.838\text{ }g\text{ }{{H}_{2}}O \\\ \end{aligned}$
- As we know that redox reactions have mainly two parts. These are oxidised half and reduced half. It is found that the oxidised half of the reaction loses electrons and hence the oxidation number of the species increases. Whereas, the reduced half of the reaction accepts electrons and hence the oxidation number of the species decreases.
- Hence, we can say that 4.838 g grams of ${{H}_{2}}O$ will be produced when 8.064 g of glucose is burned.

Note: As we know that in an oxidation reaction, the electrons are lost from an atom. An oxidizing agent is one which gains electrons.
Whereas, in reduction reaction, the electrons are gained from an atom. A reducing agent is one which loses electrons.