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. 12 g of H2
B. 2.0 g of H2
C. 3.0 g of H2
D. 10 g of H2

1 answer

To solve this problem, we need to determine how much hydrogen (H2) is required to react with the given amount of nitrogen (N2) based on the chemical equation:

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

From the balanced equation, we see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

  1. Calculate moles of nitrogen:

    • The molar mass of N2 (nitrogen) is approximately 28.02 g/mol.
    • The number of moles of nitrogen in 10.0 grams can be calculated as follows: \[ \text{moles of N}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 , \text{g}}{28.02 , \text{g/mol}} \approx 0.357 , \text{moles of N}_2 \]
  2. Calculate moles of hydrogen needed:

    • From the balanced equation, 1 mole of N2 requires 3 moles of H2. Therefore, the amount of H2 required is: \[ \text{moles of H}_2 = 3 \times \text{moles of N}_2 = 3 \times 0.357 , \text{moles} \approx 1.071 , \text{moles of H}_2 \]
  3. Convert moles of hydrogen to grams:

    • The molar mass of H2 (hydrogen) is about 2.02 g/mol.
    • The mass of H2 needed can be calculated as: \[ \text{mass of H}_2 = \text{moles of H}_2 \times \text{molar mass of H}_2 = 1.071 , \text{moles} \times 2.02 , \text{g/mol} \approx 2.16 , \text{g of H}_2 \]

Based on the calculations, the closest answer would be approximated to 2.0 grams of H2.

Thus, the answer is: B. 2.0 g of H2.