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, for now, nitrogen is balanced without adding coefficients.

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 balanced without adding coefficients.

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, the number of hydrogen atoms in $HCN$ and $H_2O$ is $1 + 2n$ (where $n$ is the coefficient of $H_2O$). Let's start by writing the un - balanced equation with placeholders: $aNH_3 + bO_2 + cCH_4
ightarrow dHCN+eH_2O$. Since $a = 1$ and $c = 1$ for nitrogen and carbon balance, the hydrogen atoms give $3 + 4=1 + 2e$, so $e = 3$.

Step4: Balance oxygen atoms

The number of oxygen atoms in $O_2$ is $2b$ and in $H_2O$ is $e$. Since $e = 3$, then $2b=3$, but we need whole - number coefficients. Multiply the entire equation by 2 to get whole - number coefficients.
The balanced equation is $2NH_3(g)+3O_2(g)+2CH_4(g)
ightarrow2HCN(aq)+6H_2O(l)$

Answer:

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