Q.Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: $N_2

Concept: Limiting Reagent & Mass Calculation

First, balance the equation:

$N_2 + 3H_2 \rightarrow 2NH_3$

Molar masses:

$N_2 = 28.0\ \text{g/mol}$, $H_2 = 2.016\ \text{g/mol}$, $NH_3 = 17.03\ \text{g/mol}$

Step 1 – Find moles of each reactant

Moles of $N_2 = \frac{2.00 \times 10^3}{28.0} = 71.43\ \text{mol}$

Moles of $H_2 = \frac{1.00 \times 10^3}{2.016} = 496.0\ \text{mol}$

Step 2 – Identify the limiting reagent

From the equation, 1 mol $N_2$ requires 3 mol $H_2$.

$71.43$ mol $N_2$ would need $71.43 \times 3 = 214.3$ mol $H_2$.

We have $496.0$ mol $H_2$ — more than enough. So $N_2$ is limiting.

Step 3 – Mass of ammonia produced

1 mol $N_2$ gives 2 mol $NH_3$.

Moles of $NH_3 = 2 \times 71.43 = 142.86\ \text{mol}$ …

Concept and Intuition

The balanced equation tells us the mole ratio in which reactants combine:

This means: 1 molecule of $N_2$ needs 3 molecules of $H_2$ to make 2 molecules of $NH_3$. In terms of moles: 1 mol $N_2$ reacts with 3 mol $H_2$ to give 2 mol $NH_3$.

When we are given masses of both reactants, we cannot simply compare grams — because different substances have different molar masses. The only fair comparison is in moles, and then we check which reactant runs out first. That reactant is the limiting reagent, and it determines how much product forms.

Step-by-step solution

1. Write the balanced equation and note molar masses.

Molar masses:

• $N_2$: $2 \times 14.0 = 28.0$ g/mol
• $H_2$: $2 \times 1.0 = 2.0$ g/mol
• $NH_3$: $14.0 + 3 \times 1.0 = 17.0$ g/mol

2. Convert given masses to moles.

For $N_2$:

For $H_2$:

3. Determine the limiting reagent.

From the equation, 1 mol $N_2$ requires 3 mol $H_2$.

So, 71.43 mol $N_2$ would require:

We have 500 mol $H_2$ — that is more than enough. So $N_2$ is the limiting reagent; it will be consumed completely.

Method: Limiting Reagent Approach (Stoichiometry)

This method identifies which reactant runs out first, then calculates product mass and leftover reactant.

Step 1: Balance the chemical equation

The given equation is unbalanced:

Balance it:

Step 2: Find molar masses

• Molar mass of $N_2$ = $2 \times 14 = 28$ g/mol
• Molar mass of $H_2$ = $2 \times 1 = 2$ g/mol
• Molar mass of $NH_3$ = $14 + 3 \times 1 = 17$ g/mol

Step 3: Convert given masses to moles

Given:

• $N_2$ mass = $2.00 \times 10^3$ g
• $H_2$ mass = $1.00 \times 10^3$ g

Moles of $N_2$:

Moles of $H_2$:

Step 4: Identify the limiting reagent

From the balanced equation:

1 mol $N_2$ requires 3 mol $H_2$

So, 71.43 mol $N_2$ would require:

We have 500 mol $H_2$ — more than enough.

Thus, $N_2$ is the limiting reagent.

Step 5: Calculate mass of ammonia produced

From the equation:

1 mol $N_2$ produces 2 mol $NH_3$

So, 71.43 mol $N_2$ produces:

Mass of $NH_3$:

$142.86 \times 17 = 2428.62 \text{ g}$ …

Here is a breakdown of the common mistakes students make on this stoichiometry problem, along with how to avoid each one.

🚨 Mistake 1: Forgetting to Balance the Equation First

The Error: Students use the given equation $N_2 + H_2 \rightarrow 2NH_3$ as-is. They assume 1 mole of $N_2$ reacts with 1 mole of $H_2$ to give 2 moles of $NH_3$.

Why it’s wrong: The equation is unbalanced. Hydrogen atoms are not conserved. The correct balanced equation is:

This means 1 mole of $N_2$ requires 3 moles of $H_2$, not 1.

How to avoid: Always check and balance the chemical equation before doing any calculation. Count atoms of each element on both sides. If they don’t match, adjust coefficients.

🚨 Mistake 2: Confusing Mass with Moles (The "Direct Mass" Trap)

The Error: Students try to compare the given masses directly (e.g., “$2000$ g of $N_2$ vs $1000$ g of $H_2$”) to decide which reactant is limiting.

Why it’s wrong: Chemical reactions happen between molecules (moles), not grams. $1$ g of $H_2$ contains many more molecules than $1$ g of $N_2$ because $H_2$ is much lighter.

How to avoid: Always convert mass to moles first using the formula:

• Molar mass of $N_2 = 28.0$ g/mol
• Molar mass of $H_2 = 2.016$ g/mol (often rounded to $2.0$ g/mol in exams)

🚨 Mistake 3: Incorrectly Identifying the Limiting Reagent

The Error: After finding moles, students assume the reactant with the smaller number of moles is the limiting reagent.

Why it’s wrong: The limiting reagent depends on the stoichiometric ratio. For example:

• Moles of $N_2 = \frac{2000}{28} \approx 71.43$ mol
• Moles of $H_2 = \frac{1000}{2} = 500$ mol

Here, $N_2$ has fewer moles, but the reaction needs 3 moles of $H_2$ for every 1 mole of $N_2$. So $71.43$ mol of $N_2$ would need $71.43 \times 3 = 214.3$ mol of $H_2$. Since we have $500$ mol of $H_2$, $N_2$ is the limiting reagent, not $H_2$.

How to avoid: Use the "divide by coefficient" method:

The reactant with the smallest result is the limiting reagent.

• For $N_2$: $71.43 \div 1 = 71.43$
• For $H_2$: $500 \div 3 \approx 166.67$
• $71.43 < 166.67$, so $N_2$ is limiting.

🚨 Mistake 4: Using the Wrong Mole Ratio for Product Calculation

The Error: After finding the limiting reagent, students use the mole ratio from the unbalanced equation or use the wrong reactant’s moles to find product moles.

Why it’s wrong: The product yield is determined only by the limiting reagent. From the balanced equation:

So, moles of $NH_3 = 2 \times \text{moles of limiting } N_2 = 2 \times 71.43 = 142.86$ mol.

How to avoid: Write the balanced equation clearly. Circle the limiting reagent. Then, set up a proportion using only that reactant’s coefficient and the product’s coefficient.

🚨 Mistake 5: Forgetting to Convert Product Moles Back to Mass

The Error: Students stop after finding moles of $NH_3$ (e.g., $142.86$ mol) and write that as the answer.

Why it’s wrong: The question asks for mass of ammonia, not moles.

How to avoid: Always check the unit asked in the question. Convert moles to mass using:

• Molar mass of $NH_3 = 14 + 3(1) = 17$ g/mol
• Mass of $NH_3 = 142.86 \times 17 \approx 2428.6$ g …