Q.50.0 kg of $N_2$

Concept: Limiting Reagent & Stoichiometry – the reactant that gets consumed first determines the maximum product.

Step 1: Write the balanced equation

Step 2: Convert masses to moles

Molar mass $N_2 = 28.0\ \text{g/mol}$, $H_2 = 2.016\ \text{g/mol}$

Step 3: Find the limiting reagent

From the equation, $1\ \text{mol } N_2$ requires $3\ \text{mol } H_2$.

For $1785.7\ \text{mol } N_2$, needed $H_2 = 1785.7 \times 3 = 5357.1\ \text{mol}$

We have only $4.96 \times 10^3\ \text{mol } H_2$ — so $H_2$ is the limiting reagent.

Step 4: Calculate $NH_3$ formed …

This problem is a classic limiting reagent calculation. The idea is simple: in a chemical reaction, reactants are consumed in a fixed mole ratio. If you have more of one reactant than the other, the one that runs out first (the limiting reagent) determines how much product you can make. Here, we have nitrogen and hydrogen gas reacting to form ammonia.

The balanced equation is:

This tells us that 1 mole of $N_2$ requires exactly 3 moles of $H_2$ to react completely. If the actual mole ratio of $H_2$ to $N_2$ is less than 3, hydrogen is limiting; if greater than 3, nitrogen is limiting.

Let’s work through the numbers step by step.

1. Convert the given masses to moles.

   Molar mass of $N_2$ = $2 \times 14.0 = 28.0 \text{ g/mol}$

   Molar mass of $H_2$ = $2 \times 1.008 = 2.016 \text{ g/mol}$ (we can use 2.0 g/mol for simplicity in many exam contexts, but let’s be precise here).

   Mass of $N_2$ = 50.0 kg = $50.0 \times 10^3$ g

   Moles of $N_2$ = $\frac{50.0 \times 10^3}{28.0} = 1785.7 \text{ mol}$ (approx.)

   Mass of $H_2$ = 10.0 kg = $10.0 \times 10^3$ g

   Moles of $H_2$ = $\frac{10.0 \times 10^3}{2.016} = 4960.3 \text{ mol}$ (approx.)

2. Determine the stoichiometric requirement.

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

   For 1785.7 mol $N_2$, the required $H_2$ = $3 \times 1785.7 = 5357.1 \text{ mol}$.

   But we only have 4960.3 mol $H_2$. That’s less than needed. So hydrogen is the limiting reagent.

   Alternatively, check the actual mole ratio:

   $\frac{\text{moles } H_2}{\text{moles } N_2} = \frac{4960.3}{1785.7} = 2.78$, which is less than 3. Confirms $H_2$ is limiting.

3. Calculate the amount of $NH_3$ formed.

   Since $H_2$ is limiting, we use its moles to find product.

   From the equation: 3 mol $H_2$ produce 2 mol $NH_3$.

   So, 1 mol $H_2$ produces $\frac{2}{3}$ mol $NH_3$.

   Moles of $NH_3$ formed = $\frac{2}{3} \times 4960.3 = 3306.9 \text{ mol}$.

4. Convert moles of $NH_3$ to mass. …

Method: Limiting Reagent Approach via Mole Comparison

This problem uses the limiting reagent method — we find which reactant runs out first by converting masses to moles and comparing the stoichiometric ratios.

Step 1: Write the balanced chemical equation

Step 2: Convert given masses to moles

• Molar mass of $N_2$ = $2 \times 14 = 28 \text{ g/mol}$

  Moles of $N_2$ = $\frac{50.0 \times 10^3 \text{ g}}{28 \text{ g/mol}} = \mathbf{1785.7 \text{ mol}}$

• Molar mass of $H_2$ = $2 \times 1.008 = 2.016 \text{ g/mol}$ (the textbook uses this precise value, not the rounded 2 g/mol, in this particular worked example)

  Moles of $H_2$ = $\frac{10.0 \times 10^3 \text{ g}}{2.016 \text{ g/mol}} = \mathbf{4.96 \times 10^3 \text{ mol}}$

Step 3: Find the limiting reagent

From the equation:

• 1 mol $N_2$ requires 3 mol $H_2$

Check with available $N_2$:

$1785.7 \text{ mol } N_2$ would need $1785.7 \times 3 = \mathbf{5357.1 \text{ mol } H_2}$

But we only have 4.96 × 10³ mol $H_2$ — not enough.

Therefore, $H_2$ is the limiting reagent.

Step 4: Calculate $NH_3$ formed

From the equation:

• 3 mol $H_2$ produces 2 mol $NH_3$ …

Common Mistakes in Molecular Mass Calculation & Limiting Reagent Problems

Students frequently lose marks on this exact type of problem. Here are the most common errors and how to avoid each:

1. ✗ Forgetting to Convert kg to g (or to moles correctly)

The Mistake:

Using 50.0 and 10.0 directly as grams, or forgetting that $1 \text{ kg} = 1000 \text{ g}$.

How to Avoid:

Always write the conversion explicitly:

Pro tip: Circle the units in the question. If you see "kg", immediately write the conversion factor.

2. ✗ Using Wrong Molar Mass Values

The Mistake:

Using $N = 14 \text{ g/mol}$ for $N_2$ (instead of $28 \text{ g/mol}$), or using $H = 1 \text{ g/mol}$ for $H_2$ (instead of $2 \text{ g/mol}$).

How to Avoid:

Remember:

• $N_2$ means two nitrogen atoms → $2 \times 14 = 28 \text{ g/mol}$
• $H_2$ means two hydrogen atoms → $2 \times 1 = 2 \text{ g/mol}$
• $NH_3$ means $14 + 3(1) = 17 \text{ g/mol}$

Write the molecular formula and count atoms before calculating.

3. ✗ Not Balancing the Chemical Equation First

The Mistake:

Using the ratio $N_2 : H_2 : NH_3$ as $1:1:1$ from the unbalanced equation $N_2 + H_2 \rightarrow NH_3$.

How to Avoid:

Always balance the equation before doing any mole calculations:

The correct mole ratio is:

4. ✗ Comparing Masses Instead of Moles for Limiting Reagent

The Mistake:

Thinking "10 kg is less than 50 kg, so $H_2$ is limiting" — this is wrong because different substances have different molar masses.

How to Avoid:

Always convert both reactants to moles first, then compare using the balanced equation ratio.

Correct approach:

5. ✗ Incorrectly Determining the Limiting Reagent

The Mistake:

Comparing raw mole numbers (1785.7 vs 5000) and concluding $N_2$ is limiting because it has fewer moles.

How to Avoid:

Use the "divide by coefficient" method:

For $N_2$: $\frac{1785.7}{1} = 1785.7$

For $H_2$: $\frac{4960}{3} = 1653.3$

The reactant with the smaller quotient is limiting. …