Question

Complete the following reaction and balance it:

${\mathrm{Al}}_2{\mathrm{O}}_3+\mathrm{NaOH}\to$

Answer 1

The reaction can be solved in two steps:

1. Complete the reaction by writing the products.
2. Balance the equation.

1. Complete the reaction by writing the products:

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3+\mathrm{NaOH}\to {\mathrm{NaAlO}}_2+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Aluminium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{aluminate})\left(\mathrm{Water}\right) \end{gathered}$$

2. Balancing of the reaction by hit and trial method:

• A chemical equation is said to be balanced when atoms of every element in the chemical equation are equal in number on both (reactant and product) sides.
• To balance the chemical equation, begin by multiplying the element or compound with fewer atoms with a number that would make the number equal on both sides.

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3+\mathrm{NaOH}\to {\mathrm{NaAlO}}_2+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Aluminium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{aluminate})\left(\mathrm{Water}\right) \end{gathered}$$

Left hand side (LHS) is the reactant side and Right hand side (RHS) is the product side.

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Aluminium (Al)	2	1
Oxygen (O)	4	3
Sodium (Na)	1	1
Hydrogen (H)	1	2

Balance the equation by multiplying Sodium aluminate (NaAlO₂) on RHS with 2 as the number of atoms of Sodium and Aluminium are less on RHS.

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3+\mathrm{NaOH}\to 2{\mathrm{NaAlO}}_2+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Aluminium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{aluminate})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Aluminium (Al)	2	2
Oxygen (O)	4	5
Sodium (Na)	1	2
Hydrogen (H)	1	2

As the number of atoms is still unequal on both sides, multiply Sodium hydroxide (NaOH) with 2.

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3+2\mathrm{NaOH}\to 2{\mathrm{NaAlO}}_2+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Aluminium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{aluminate})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Aluminium (Al)	2	2
Oxygen (O)	5	5
Sodium (Na)	2	2
Hydrogen (H)	2	2

As the number of all the atoms is equal on both sides, the chemical equation is now balanced.

So, the overall complete and balanced chemical equation is

$$\begin{gathered} {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)\to 2{\mathrm{NaAlO}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Aluminium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{aluminate})\left(\mathrm{Water}\right) \end{gathered}$$