Question

Write the balanced chemical equation for Aluminium nitrate +sodium hydroxide to form aluminium hydroxide and sodium nitrate?

Answer 1

We need to know that when the number of atoms on the reactant side is equal to the number of atoms on the product side. A balanced chemical equation follows the Law of Conservation of mass.
The Law of Conservation states that mass can neither be created nor be destroyed.

Complete answer:
We also remember that balancing a chemical equation means making the atoms similar on both sides of the reaction.
First we convert the word equation into symbol equation. For that we wrote the formula for the chemical compound.
The formula for Aluminium nitrate is $Al{\left( {N{O_3}} \right)_3}$
For sodium hydroxide it is $NaOH$
For aluminium hydroxide it is $Al{\left( {OH} \right)_3}$
For sodium nitrate it is $NaN{O_3}$
Write the equation
$Al{\left( {N{O_3}} \right)_{3\;}} + NaOH\; \to \;Al{\left( {OH} \right)_3}\; + NaN{O_3}$
Now count the number of atoms of each type on reactant side excluding the and then on product
For writing the number of atoms in case of ${\left( {N{O_3}} \right)_3}$ we have to consider the subscript also so number of atoms for nitrogen on reactant side becomes $3$ and that of oxygen becomes $9$whereas $1$ atom is present in $NaOH$ also which gets added. For the product side for oxygen it becomes $3 + 3 = 6$ , $3$ from $Al{\left( {OH} \right)_3}$ and $3$ $NaN{O_3}$.
So total atoms on reactant side becomes
Nitrogen: $3$
Oxygen: $9 + 1 = 10$
Hydrogen: $1$
Aluminium: $1$
On the product side it becomes
Nitrogen: $1$
Oxygen: $3 + 3 = 6$
Hydrogen: $3$
Aluminium: $1$

Atoms	Number on L.H.S	Number on R.H.S.
$Al$	$1$	$1$
$N$	$3$	$1$
$O$	$10$	$6$
$H$	$1$	$3$
$Na$	$1$	$1$

Now to balance the number of nitrogen atoms on R.H.S. we multiply by 3 it becomes
$Al{\left( {N{O_3}} \right)_{3\;}} + NaOH \to Al{\left( {OH} \right)_3} + 3NaN{O_3}$
The number of atoms becomes,

Atoms	Number on L.H.S	Number on R.H.S.
$Al$	$1$	$1$
$N$	$3$	$3$
$O$	$10$	$12$
$H$	$1$	$3$
$Na$	$1$	$3$

Now we balance the number of Na atoms on L.H.S. by multiplying by 3
$Al{\left( {N{O_3}} \right)_{3\;}} + 3NaOH \to Al{\left( {OH} \right)_3} + 3NaN{O_3}$
The number of atoms becomes,

Atoms	Number on L.H.S	Number on R.H.S.
$Al$	$1$	$1$
$N$	$3$	$3$
$O$	$12$	$12$
$H$	$3$	$3$
$Na$	$3$	$3$

The number of atoms on both sides are the same hence, the equation is balanced.
The balanced chemical equation is:
$Al{\left( {N{O_3}} \right)_{3\;}} + 3NaOH \to Al{\left( {OH} \right)_3} + 3NaN{O_3}$

Note:
We also remember that balancing a chemical enables us to solve and calculate the number of moles of compound required for a reaction without solving it.