Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. \\(\ce{n_{2}(g) + h_{2}(g) -> nh_{3}(g)}\\)

Explanation:

Step1: Balance Nitrogen atoms

On the left, we have \( N_2 \), so 2 N atoms. On the right, \( NH_3 \) has 1 N atom. To balance N, we put a coefficient of 2 in front of \( NH_3 \). Now the equation is \( N_2(g) + H_2(g)
ightarrow 2NH_3(g) \).

Step2: Balance Hydrogen atoms

Now, on the right, \( 2NH_3 \) has \( 2\times3 = 6 \) H atoms. On the left, \( H_2 \) has 2 H atoms. So we need a coefficient of 3 in front of \( H_2 \) to get 6 H atoms. The balanced equation is \( N_2(g) + 3H_2(g)
ightarrow 2NH_3(g) \).

Answer:

\( \boldsymbol{N_2(g) + 3H_2(g)
ightarrow 2NH_3(g)} \)