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

\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

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General Chemistry FAQs

How do I balance \( \mathrm{N_2 + H_2 \rightarrow NH_3} \)?

Balance atoms. Let coefficients be \(a\), \(b\), \(c\): \(a\mathrm{N_2}+b\mathrm{H_2}\rightarrow c\mathrm{NH_3}\). N: \(2a=c\). H: \(2b=3c\). Smallest integers: \(a=1\), \(b=3\), \(c=2\). So \( \mathrm{N_2+3H_2\rightarrow 2NH_3}\).

Why do the coefficients \(1:3:2\) work for \( \mathrm{N_2 + 3H_2 \rightarrow 2NH_3} \)?

Check atoms. Left: N \(=2\cdot 1=2\). H \(=2\cdot 3=6\). Right: N \(=2\cdot 1=2\). H \(=2\cdot 3=6\). Both elements match.

What is the stoichiometric ratio between \( \mathrm{H_2} \) and \( \mathrm{NH_3} \)?

From \( \mathrm{N_2+3H_2\rightarrow 2NH_3} \), the mole ratio is \( \mathrm{H_2:NH_3}=3:2 \). Meaning \(3\) moles \( \mathrm{H_2} \) produce \(2\) moles \( \mathrm{NH_3} \) (assuming complete reaction).

If I start with \(5\) mol \( \mathrm{N_2} \), how many moles \( \mathrm{NH_3} \) form?

Use the ratio \( \mathrm{N_2:NH_3}=1:2 \). So \(5\) mol \( \mathrm{N_2} \rightarrow 10\) mol \( \mathrm{NH_3} \). (Assumes excess \( \mathrm{H_2} \) and complete reaction.)

If \(10\) mol \( \mathrm{H_2} \) are available, what is the maximum \( \mathrm{NH_3} \) formed?

Ratio \( \mathrm{H_2:NH_3}=3:2 \). So moles \( \mathrm{NH_3} =10\cdot \frac{2}{3}=6.67\) mol. Check that \( \mathrm{N_2} \) needed is \(10\cdot\frac{1}{3}=3.33\) mol.

How do I find the limiting reactant in \( \mathrm{N_2 + 3H_2 \rightarrow 2NH_3} \)?

Compare based on stoichiometric requirements. Compute required \( \mathrm{H_2} \) for given \( \mathrm{N_2} \) using \(3\) mol \( \mathrm{H_2} \) per \(1\) mol \( \mathrm{N_2} \). If available \( \mathrm{H_2} \) is less than required, \( \mathrm{H_2} \) limits; otherwise \( \mathrm{N_2} \) limits.
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