How do you write a balanced equation for combustion (reaction with oxygen gas) of glucose, ${{C}_{6}}{{H}_{12}}{{O}_{6}}$to give $C{{O}_{2}}and\,{{H}_{2}}O$?

Hint: We can balance the given reaction by balancing the individual atoms involved in it. For this we will have to make the number of individual atoms equal in LHS and RHS. Firstly, we will write the skeletal equation and then balance the $C$ atom followed by $H,O$ involved in the reaction.

Complete step-by-step answer:Step 1: The skeletal chemical reaction can be written as;
${{C}_{6}}{{H}_{12}}{{O}_{6}}\,+\,{{O}_{2}}\,\to \,C{{O}_{2}}\,+\,{{H}_{2}}O$
Step 2: Balance the Carbon atoms.
 In the above reaction we see that the $C$ atoms on the LHS are $6$ and on the RHS only $1$ carbon atoms are present. So, to balance the carbon we will add $6$in front of $C{{O}_{2}}$. The given reaction will become;
${{C}_{6}}{{H}_{12}}{{O}_{6}}\,+\,{{O}_{2}}\to \,6C{{O}_{2}}\,+\,{{H}_{2}}O$…. $\left( i \right)$
Step 3: Balance the H atoms.
In the reaction $\left( i \right)$ we see that there are $12\,H$ atoms on the LHS and $2\,H$ atoms on the RHS. So, to balance hydrogen we will add $6$in front of ${{H}_{2}}O$. The reaction will become
${{C}_{6}}{{H}_{12}}{{O}_{6}}\,+{{O}_{2}}\to \,6C{{O}_{2}}\,+\,6{{H}_{2}}O$ …$\left( ii \right)$
Step 4: Balance the $O$atoms.
In the reaction $\left( ii \right)$ we see that there are $18\,O$ atoms on the RHS and only $8\,O$ atoms on the LHS. So, to balance them we will add $6$in front of oxygen as shown,
${{C}_{6}}{{H}_{12}}{{O}_{6}}\,+6C{{O}_{2}}\,\to \,6C{{O}_{2}}\,+\,6{{H}_{2}}O$
The above reaction is this balanced reaction for the combustion of glucose.
It has an equal number of carbon hydrogen and oxygen atoms on LHS and RHS.

Note:Always count whether the number of atoms on LHS and RHS of the reaction are equal or not. If they are equal then the reaction will be properly balanced. If ever any fraction terms appear while balancing, convert it into the whole number by multiplying by a suitable number.