Question

Write the balanced chemical equations for the following reactions

Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia.

Answer 1

Balanced chemical equation:

A chemical equation in which the number of atoms of each element on both sides of the equation is equal is called a balanced chemical equation.

We can balance the following equation by using the hit-trial method.

Step-1: Representation of chemical equation:

When Magnesium nitride reacts with water, magnesium hydroxide and ammonia are formed.

$\underset{(\mathrm{Magnesium}\mathrm{nitride})}{{\mathrm{Mg}}_3{\mathrm{N}}_2}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_2\mathrm{O}}⟶\underset{(\mathrm{magnesium}\mathrm{hydroxide})}{\mathrm{Mg}(\mathrm{OH}{)}_2}+\underset{\left(\mathrm{ammonia}\right)}{{\mathrm{NH}}_3}$

Step-2: Calculating atoms on both sides of equation.

Elements	The number of atoms on the reactant side	The number of atoms on the product side
$\mathrm{Mg}$	$3$	$1$
$\mathrm{N}$	$2$	$1$
$\mathrm{H}$	$2$	$5$
$\mathrm{O}$	$1$	$2$

Step-3: Balancing atoms on both sides of the equation:

To balance magnesium $\left(\mathrm{Mg}\right)$ on both sides, we will add '3' against $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2$ on product side.

${\mathrm{Mg}}_3{\mathrm{N}}_2+{\mathrm{H}}_2\mathrm{O}⟶3\mathrm{Mg}{\left(\mathrm{OH}\right)}_2+{\mathrm{NH}}_3$

Elements	The number of atoms on the reactant side	The number of atoms on the product side
Mg	3	3
N	2	1
H	2	8
O	1	6

Now to balance $\left(\mathrm{N}\right)$ on both sides, we will add ‘2’ against $\left({\mathrm{NH}}_3\right)$ on the product side.

${\mathrm{Mg}}_3{\mathrm{N}}_2+{\mathrm{H}}_2\mathrm{O}⟶3\mathrm{Mg}{\left(\mathrm{OH}\right)}_2+2{\mathrm{NH}}_3$

Elements	The number of atoms on the reactant side	The number of atoms on the product side
Mg	3	3
N	2	2
H	2	12
O	1	6

Now to balance oxygen and hydrogen on both sides, we will add ‘6’ against $\left({\mathrm{H}}_2\mathrm{O}\right)$ on reactant side.

$\underset{(\mathrm{Magnesium}\mathrm{nitride})}{{\mathrm{Mg}}_3{\mathrm{N}}_2}+\underset{\left(\mathrm{Water}\right)}{6{\mathrm{H}}_2\mathrm{O}}⟶\underset{(\mathrm{magnesium}\mathrm{hydroxide})}{3\mathrm{Mg}(\mathrm{OH}{)}_2}+\underset{\left(\mathrm{ammonia}\right)}{2{\mathrm{NH}}_3}$

Elements	The number of atoms on the reactant side	The number of atoms on the product side
Mg	3	3
N	2	2
H	12	12
O	6	6

Now, we can see that atoms are the same on both sides of the reaction.

Hence, the equation is balanced.

$\underset{(\mathrm{Magnesium}\mathrm{nitride})}{{\mathrm{Mg}}_3{\mathrm{N}}_2\left(\mathrm{s}\right)}+\underset{\left(\mathrm{Water}\right)}{6{\mathrm{H}}_2\mathrm{O}}\left(\mathrm{l}\right)⟶\underset{(\mathrm{magnesium}\mathrm{hydroxide})}{3\mathrm{Mg}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)}+\underset{\left(\mathrm{ammonia}\right)}{2{\mathrm{NH}}_3\left(\mathrm{g}\right)}$