Question

The chemically balanced equation presented below illustrates the reaction between nitrogen (N2) and hydrogen (H2) resulting in the formation of ammonia (NH3).
N2 + 3H2 → 2NH3
In the event that 10.0 grams of nitrogen participate in a reaction to produce 12.0 grams of ammonia, what would be the quantity of hydrogen, in grams, have been during this reaction?
A. 3.0 g of H2
B. 10 g of H2
C. 12 g of H2
D. 2.0 g of H2

Answer 1

To determine the quantity of hydrogen used in the reaction, we can use the balanced chemical equation:

\[
N_2 + 3H_2 \rightarrow 2NH_3
\]

According to the stoichiometry of the reaction, 1 mole of \(N_2\) reacts with 3 moles of \(H_2\) to produce 2 moles of \(NH_3\).

**Molecular weights:**
- Molecular weight of \(N_2\) = 28.0 g/mol (14.0 g/mol for each N)
- Molecular weight of \(H_2\) = 2.0 g/mol (1.0 g/mol for each H)
- Molecular weight of \(NH_3\) = 17.0 g/mol (14.0 g/mol for N + 3.0 g/mol for H)

**Steps to calculate:**

1. **Convert grams of \(N_2\) to moles:**
\[
\text{Moles of } N_2 = \frac{10.0 \text{ g}}{28.0 \text{ g/mol}} \approx 0.357 \text{ moles}
\]

2. **Use stoichiometry to find moles of \(H_2\):**
From the balanced equation, 1 mole of \(N_2\) requires 3 moles of \(H_2\). Thus, the moles of \(H_2\) needed for 0.357 moles of \(N_2\) is:
\[
\text{Moles of } H_2 = 0.357 \text{ moles } N_2 \times \frac{3 \text{ moles } H_2}{1 \text{ mole } N_2} = 1.071 \text{ moles } H_2
\]

3. **Convert moles of \(H_2\) to grams:**
\[
\text{Grams of } H_2 = 1.071 \text{ moles} \times 2.0 \text{ g/mol} = 2.142 \text{ g}
\]

Approximately, we can round this to 2.0 grams for practical purposes.

**Answer:**
D. 2.0 g of \(H_2\)