Q. We compute \(n^2 + 3h^2 \to 2nh^3\).
Answer
Balance the atoms in
\[ n \text{N}_2 + h \text{H}_2 \rightarrow 2 \text{N}h_3 \]
Because the product has \(2\) molecules of \( \text{N}h_3 \), it contains \(2\) nitrogen atoms and \(6\) hydrogen atoms. Since \( \text{N}_2 \) has \(2\) nitrogen atoms per molecule, we need \(n = 1\). Since \( \text{H}_2 \) has \(2\) hydrogen atoms per molecule, we need \(h = 3\).
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
Detailed Explanation
We are given the chemical reaction:
Unbalanced equation:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
We must determine whether this equation is balanced, and (if needed) balance it.
## Step 1: Interpret the reactants and products
The reactants are:
- \(\text{N}_2\): nitrogen molecules
- \(3\text{H}_2\): three hydrogen molecules worth of hydrogen
The product is:
- \(2\text{NH}_3\): two ammonia molecules
## Step 2: Count nitrogen atoms
On the left (reactant side):
- \(\text{N}_2\) contains 2 nitrogen atoms
So total nitrogen on the left is:
\[ 2 \]
On the right (product side):
- \(\text{NH}_3\) contains 1 nitrogen atom per molecule
- There are \(2\text{NH}_3\) molecules, so total nitrogen is \(2 \times 1 = 2\)
So nitrogen on the right is:
\[ 2 \]
✅ Nitrogen atoms balance.
## Step 3: Count hydrogen atoms
On the left (reactant side):
- \(\text{H}_2\) contains 2 hydrogen atoms per molecule
- There are \(3\text{H}_2\) molecules, so total hydrogen is \(3 \times 2 = 6\)
So total hydrogen on the left is:
\[ 6 \]
On the right (product side):
- \(\text{NH}_3\) contains 3 hydrogen atoms per molecule
- There are \(2\text{NH}_3\) molecules, so total hydrogen is \(2 \times 3 = 6\)
So hydrogen on the right is:
\[ 6 \]
✅ Hydrogen atoms balance.
## Step 4: Conclude whether the equation is balanced
Since both nitrogen and hydrogen atoms match on both sides, the given equation is already balanced.
## Final balanced equation
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
General Chemistry FAQs
How do I balance \( \mathrm{N_2 + H_2 \rightarrow NH_3} \)?
Why do the coefficients \(1:3:2\) work for \( \mathrm{N_2 + 3H_2 \rightarrow 2NH_3} \)?
What is the stoichiometric ratio between \( \mathrm{H_2} \) and \( \mathrm{NH_3} \)?
If I start with \(5\) mol \( \mathrm{N_2} \), how many moles \( \mathrm{NH_3} \) form?
If \(10\) mol \( \mathrm{H_2} \) are available, what is the maximum \( \mathrm{NH_3} \) formed?
How do I find the limiting reactant in \( \mathrm{N_2 + 3H_2 \rightarrow 2NH_3} \)?
Practice with helpful NH3 math.
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