The chemically balanced equation presented below illustrates the reaction between nitrogen (N2) and hydrogen (H2) resulting in the formation of ammonia (NH3)

Question

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?

Answer 1

To determine the quantity of hydrogen consumed in the reaction, we need to use stoichiometry based on the balanced equation:

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

1. Calculate moles of nitrogen (N2):

   • The molar mass of N2 (Nitrogen) is approximately \( 28.02 , \text{g/mol} \) (14.01 g/mol per N atom).
   • Moles of N2 from 10.0 grams: \[ \text{Moles of N}_2 = \frac{10.0 , \text{g}}{28.02 , \text{g/mol}} \approx 0.356 , \text{mol} \]

2. Use the stoichiometry of the reaction to find moles of hydrogen (H2) needed:

   • According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2.
   • Therefore, moles of H2 needed for 0.356 moles of N2: \[ \text{Moles of H}_2 = 3 \times \text{Moles of N}_2 = 3 \times 0.356 \approx 1.068 , \text{mol} \]

3. Calculate grams of hydrogen (H2) from moles:

   • The molar mass of H2 (Hydrogen) is approximately \( 2.02 , \text{g/mol} \).
   • Grams of H2: \[ \text{Mass of H}_2 = \text{Moles of H}_2 \times \text{Molar mass of H}_2 = 1.068 , \text{mol} \times 2.02 , \text{g/mol} \approx 2.16 , \text{g} \]

Thus, the quantity of hydrogen that was consumed during this reaction is approximately 2.16 grams.