Question

Balance the following chemical equation:

$\mathrm{BeO}+\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{BeO}}_2+{\mathrm{H}}_2\mathrm{O}$

Answer 1

Balancing of the reaction by trial and error 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 which has less number of atoms with a number that would make the number equal on both sides.

$$\begin{gathered} \mathrm{BeO}+\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{BeO}}_2+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Berrylium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{beryllate})\left(\mathrm{Water}\right) \end{gathered}$$

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.
Beryllium (Be)	1	1
Oxygen (O)	2	3
Sodium (Na)	1	2
Hydrogen (H)	1	2

Balance the equation by multiplying Sodium hydroxide (NaOH) on LHS with 2 as the number of atoms of Sodium are less on LHS.

$$\begin{gathered} \mathrm{BeO}+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{BeO}}_2+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Berrylium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{beryllate})\left(\mathrm{Water}\right) \end{gathered}$$

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

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

So, the overall complete and balanced chemical equation is

$$\begin{gathered} \mathrm{BeO}\left(\mathrm{s}\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{BeO}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Berrylium}\mathrm{oxide})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{beryllate})\left(\mathrm{Water}\right) \end{gathered}$$