Question

What should be the value of x (coefficient of $H_{2}O$)?

$a{C}_6{H}_{12}{O}_6+b{O}_2\to cC{O}_2+x{H}_{2}O$

• A. 4
• B. 6
• C. 12
• D. 8

Answer 1

The correct option is B

6

To obtain the value of x, both the reactants and products have to be balanced in the given equation. We start the balancing with writing the number of atoms present on reactant and the products side:

Number of Carbon atom on RHS - 6, Number of Carbon atom on LHS - 1

Number of the hydrogen atom on RHS - 12, Number of the hydrogen atom on LHS - 2

Number of oxygen atom on RHS - 8, Number of oxygen atom on RHS - 3

Now we will start balancing the carbon atoms, so on the LHS carbon atom are 6 but on RHS it is only 1. So we will multiply 6 with $C{O}_2$. Now the number of carbon atom on the RHS has become 6.
$C_6{H}_{12}{O}_6+O_2\to 6C{O}_2+H_{2}O$

After carbon, we will balance hydrogen. On the LHS, the number of hydrogen atoms is 12, but on the RHS it is only 2. So to balance the hydrogen atoms, we will multiply 6 to $H_{2}O$, the number of hydrogen atom becomes 12.

$C_6{H}_{12}{O}_6+O_2\to 6C{O}_2+6H_{2}O$

The number of oxygen atom on the RHS is 18 atoms, but on the LHS it is 8 atoms. To balance the oxygen atom, we will multiply 6 to $O_2$, the total number of oxygen atoms on the LHS is now 18 and hence they are balanced.

$C_6{H}_{12}{O}_6+6O_2\to 6C{O}_2+6H_{2}O$

Hence the equation is balanced.