11.2 State symbols  (Page 32/2)

The reactants are zinc ( $\mathrm{Zn}$ ) and hydrochloric acid ( $\mathrm{HCl}$ ). The products are zinc chloride ( $\mathrm{ZnCl}{}_2$ ) and hydrogen ( $\mathrm{H}{}_2$ ).

You will notice that the zinc atoms balance but the chlorine and hydrogen atoms don't. Since there are two chlorine atoms on the right and only one on the left, we will give $\mathrm{HCl}$ a coefficient of 2 so that there will be two chlorine atoms on each side of the equation.

When you look at the equation again, you will see that all the atoms are now balanced.

In the initial description, you were told that zinc was a metal, hydrochloric acid and zinc chloride were in aqueous solutions and hydrogen was a gas.

With a complex equation, it is always best to start with atoms that appear only once on each side i.e. $\mathrm{Na}$ , $\mathrm{N}$ and $\mathrm{S}$ atoms. Since the $\mathrm{S}$ atoms already balance, we will start with $\mathrm{Na}$ and $\mathrm{N}$ atoms. There are two $\mathrm{Na}$ atoms on the right and one on the left. We will add a second $\mathrm{Na}$ atom by giving $\mathrm{NaOH}$ a coefficient of two. There are two $\mathrm{N}$ atoms on the left and one on the right. To balance the $\mathrm{N}$ atoms, $\mathrm{NH}{}_3$ will be given a coefficient of two. The equation now looks as follows:

$\mathrm{N}$ , $\mathrm{Na}$ and $\mathrm{S}$ atoms balance, but $\mathrm{O}$ and $\mathrm{H}$ atoms do not. There are six $\mathrm{O}$ atoms and ten $\mathrm{H}$ atoms on the left, and five $\mathrm{O}$ atoms and eight $\mathrm{H}$ atoms on the right. We need to add one $\mathrm{O}$ atom and two $\mathrm{H}$ atoms on the right to balance the equation. This is done by adding another $\mathrm{H}{}_2\mathrm{O}$ molecule on the right hand side. We now need to check the equation again:

The equation is now balanced.