Question

balance the chemical equation below using the smallest possible whole - number stoichiometric coefficients.
nh₃(g) + o₂(g) + ch₄(g) → hcn(aq) + h₂o(l)

Explanation:

Step1: Balance nitrogen atoms

There is 1 nitrogen atom on the left - hand side in $NH_3$ and 1 in $HCN$ on the right - hand side, so nitrogen is already balanced for now.

Step2: Balance carbon atoms

There is 1 carbon atom in $CH_4$ on the left - hand side and 1 in $HCN$ on the right - hand side, so carbon is already balanced for now.

Step3: Balance hydrogen atoms

On the left - hand side, the total number of hydrogen atoms from $NH_3$ and $CH_4$ is $3 + 4=7$. On the right - hand side, hydrogen is in $HCN$ and $H_2O$. Let's start by adjusting the coefficients.
Let the coefficients of $NH_3$, $O_2$, $CH_4$, $HCN$ and $H_2O$ be $a$, $b$, $c$, $d$ and $e$ respectively. So the equation is $aNH_3 + bO_2 + cCH_4
ightarrow dHCN+eH_2O$.
From nitrogen balance: $a = d$. From carbon balance: $c = d$.
The hydrogen atoms: $3a + 4c=d + 2e$. Substituting $a = d$ and $c = d$ gives $3d+4d=d + 2e$, or $6d = 2e$, or $e = 3d$.
Let $d = 2$. Then $a = 2$, $c = 2$, $e = 6$.

Step4: Balance oxygen atoms

The number of oxygen atoms on the right - hand side in $H_2O$ is $e = 6$. On the left - hand side, oxygen is in $O_2$. If $e = 6$, then from the oxygen - atom balance in the equation $2b=e$, when $e = 6$, $b = 3$.

The balanced equation is $2NH_3(g)+3O_2(g)+2CH_4(g)
ightarrow2HCN(aq)+6H_2O(l)$

Answer:

$2NH_3(g)+3O_2(g)+2CH_4(g)
ightarrow2HCN(aq)+6H_2O(l)$