0.7 Representing chemical change  (Page 4/6)

Reactants: $\mathrm{C}=1$ ; $\mathrm{H}=4$ and $\mathrm{O}=2$

Products: $\mathrm{C}=1$ ; $\mathrm{H}=2$ and $\mathrm{O}=3$

If we add a coefficient of 2 to $\mathrm{H}{}_2\mathrm{O}$ , then the number of hydrogen atoms in the reactants will be 4, which is the same as for the reactants. The equation will be:

Reactants: $\mathrm{C}=1$ ; $\mathrm{H}=4$ and $\mathrm{O}=2$

Products: $\mathrm{C}=1$ ; $\mathrm{H}=4$ and $\mathrm{O}=4$

You will see that, although the number of hydrogen atoms now balances, there are more oxygen atoms in the products. You now need to repeat the previous step. If we put a coefficient of 2 in front of $\mathrm{O}{}_2$ , then we will increase the number of oxygen atoms in the reactants by 2. The new equation is:

When we check the number of atoms again, we find that the number of atoms of each element in the reactants is the same as the number in the products. The equation is now balanced.

The reactants are nitrogen ( $\mathrm{N}{}_2$ ) and hydrogen ( $\mathrm{H}{}_2$ ) and the product is ammonia ( $\mathrm{NH}{}_3$ ).

The equation is as follows:

Reactants: $\mathrm{N}=2$ and $\mathrm{H}=2$

Products: $\mathrm{N}=1$ and $\mathrm{H}=3$

In order to balance the number of nitrogen atoms, we could rewrite the equation as:

In the above equation, the nitrogen atoms now balance, but the hydrogen atoms don't (there are 2 hydrogen atoms in the reactants and 6 in the product). If we put a coefficient of 3 in front of the hydrogen ( ${\mathrm{H}}_2$ ), then the hydrogen atoms and the nitrogen atoms balance. The final equation is:

Reactants: sugar ( ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ ) and oxygen ( ${\mathrm{O}}_2$ )

Products: carbon dioxide ( ${\mathrm{CO}}_2$ ) and water ( $\mathrm{H}{}_2\mathrm{O}$ )

Reactants: $\mathrm{C}=6$ ; $\mathrm{H}=12$ and $\mathrm{O}=8$

Products: $\mathrm{C}=1$ ; $\mathrm{H}=2$ and $\mathrm{O}=3$

It is easier to start with carbon as it only appears once on each side. If we add a 6 in front of ${\mathrm{CO}}_2$ , the equation looks like this:

Reactants: $\mathrm{C}=6$ ; $\mathrm{H}=12$ and $\mathrm{O}=8$

Products: $\mathrm{C}=6$ ; $\mathrm{H}=2$ and $\mathrm{O}=13$

Let's try to get the number of hydrogens the same this time.

Reactants: $\mathrm{C}=6$ ; $\mathrm{H}=12$ and $\mathrm{O}=8$

Products: $\mathrm{C}=6$ ; $\mathrm{H}=12$ and $\mathrm{O}=18$

Reactants: $\mathrm{C}=6$ ; $\mathrm{H}=12$ and $\mathrm{O}=18$

Products: $\mathrm{C}=6$ ; $\mathrm{H}=12$ and $\mathrm{O}=18$