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?
A. 12 g of H2
B. 2.9 g of H2
C. 3.0 g of H2
D. 10 g of H2

Answer 1

To find the quantity of hydrogen (H2) that participated in the reaction, we can use the mole concept and stoichiometry based on the balanced equation:

$$N_2+3H_2\to 2N{H}_3$$

First, we need to determine the number of moles of nitrogen (N2) that reacted. The molar mass of nitrogen (N2) is approximately 28.02 g/mol (since each nitrogen atom has a molar mass of about 14.01 g).

1. Calculate the moles of nitrogen (N2): $$\text{Moles of }{N}_2=\frac{\text{mass of }{N}_2}{\text{molar mass of }{N}_2}=\frac{10.0,\text{g}}{28.02,\text{g/mol}}\approx 0.357,\text{mol},N_2$$

2. Using the balanced equation, determine the moles of hydrogen (H2) required: According to the equation, 1 mole of N2 reacts with 3 moles of H2. Therefore, the moles of H2 needed can be calculated as follows: $$\text{Moles of }{H}_2=3\times \text{moles of }{N}_2=3\times 0.357,\text{mol}\approx 1.071,\text{mol},H_2$$

3. Now, calculate the mass of hydrogen (H2) that corresponds to these moles: The molar mass of hydrogen (H2) is approximately 2.02 g/mol. $$\text{Mass of }{H}_2=\text{moles of }{H}_2\times \text{molar mass of }{H}_2=1.071,\text{mol}\times 2.02,\text{g/mol}\approx 2.16,\text{g},H_2$$

However, since the product is 12.0 g of NH3, we can also verify how much H2 is actually consumed by using the moles of ammonia produced.

4. Calculate moles of ammonia produced: The molar mass of ammonia (NH3) is approximately 17.03 g/mol: $$\text{Moles of }N{H}_3=\frac{12.0,\text{g}}{17.03,\text{g/mol}}\approx 0.705,\text{mol},N{H}_3$$

5. Using the balanced equation, determine the moles of H2 that correspond to the produced NH3: Since 2 moles of NH3 are produced from 3 moles of H2: $$\text{Moles of }{H}_2\text{ required}=3\times \frac{0.705}{2}=1.0575,\text{mol},H_2$$

6. Calculate the mass of hydrogen (H2) again: $$\text{Mass of }{H}_2=1.0575,\text{mol}\times 2.02,\text{g/mol}\approx 2.14,\text{g},H_2$$

It appears that there may be inconsistencies in the existing choices since the calculated value aligns with none of them. However, based on reasoning with the stoichiometry derived, approximately 2.14 g of hydrogen would have been consumed in the reaction, reflecting its use in producing the stated mass of ammonia.

Given the provided answer choices: A. 12 g of H2 B. 2.9 g of H2 C. 3.0 g of H2 D. 10 g of H2

The answer would be closest to B. 2.9 g of H2.