Question

__ n₂ + h₂ → __ nh₃
__ al(oh)₃ + hbr → albr₃ + __ h₂o

Explanation:

Step1: Balance the first equation ($\ce{N_{2} + H_{2} -> NH_{3}}$)

• Count N atoms: Left has 2, right has 1. So put 2 in front of $\ce{NH_{3}}$: $\ce{N_{2} + H_{2} -> 2NH_{3}}$.
• Now count H atoms: Right has $2\times3 = 6$, left has 2. So put 3 in front of $\ce{H_{2}}$: $\ce{N_{2} + 3H_{2} -> 2NH_{3}}$. Check: N (2=2), H (6=6). Balanced.

Step2: Balance the second equation ($\ce{Al(OH)_{3} + HBr -> AlBr_{3} + H_{2}O}$)

• Count Al atoms: Left 1, right 1. Good.
• Count Br atoms: Right has 3, left has 1. Put 3 in front of $\ce{HBr}$: $\ce{Al(OH)_{3} + 3HBr -> AlBr_{3} + H_{2}O}$.
• Count O atoms: Left has 3 (from $\ce{Al(OH)_{3}}$), right has 1 (from $\ce{H_{2}O}$). Put 3 in front of $\ce{H_{2}O}$: $\ce{Al(OH)_{3} + 3HBr -> AlBr_{3} + 3H_{2}O}$.
• Check H atoms: Left: $3\times1 + 3\times1 = 6$; Right: $3\times2 = 6$. Balanced.

Answer:

First equation: $\boldsymbol{1}\ce{N_{2} + 3H_{2} -> 2NH_{3}}$
Second equation: $\boldsymbol{1}\ce{Al(OH)_{3} + 3HBr -> 1AlBr_{3} + 3H_{2}O}$