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?
(1 point)
Responses
3.0 g of H2
3.0 g of H 2
12 g of H2
12 g of H 2
2.0 g of H2
2.0 g of H 2
10 g of H2
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?
(1 point)
Responses
3.0 g of H2
3.0 g of H 2
12 g of H2
12 g of H 2
2.0 g of H2
2.0 g of H 2
10 g of H2
Answers
GPT-4o mini
To find the quantity of hydrogen (H2) that participated in the reaction, we can start by using stoichiometry based on the balanced equation provided:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
From the equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
### Step 1: Calculate the moles of ammonia produced
First, we need to calculate the moles of ammonia produced (NH3). The molar mass of ammonia (NH3) is approximately:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol × 3 = 3.03 g/mol
- Total: 14.01 + 3.03 = 17.04 g/mol
Using the mass of ammonia produced (12.0 g):
\[
\text{Moles of NH}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0 \, \text{g}}{17.04 \, \text{g/mol}} \approx 0.705 \, \text{mol}
\]
### Step 2: Use the mole ratio to find moles of hydrogen
According to the equation, 2 moles of NH3 are produced from 3 moles of H2:
\[
\frac{3 \, \text{moles H}_2}{2 \, \text{moles NH}_3}
\]
Using this ratio, we can determine the moles of H2 used:
\[
\text{Moles of H}_2 = 0.705 \, \text{mol NH}_3 \times \frac{3 \, \text{mol H}_2}{2 \, \text{mol NH}_3} = 1.0575 \, \text{mol H}_2
\]
### Step 3: Calculate the mass of hydrogen used
The molar mass of hydrogen (H2) is:
- Hydrogen (H): 1.01 g/mol × 2 = 2.02 g/mol
Now we can calculate the mass of H2:
\[
\text{Mass of H}_2 = \text{moles} \times \text{molar mass} = 1.0575 \, \text{mol} \times 2.02 \, \text{g/mol} \approx 2.13 \, \text{g}
\]
Based on the options provided, this closest matches one of the responses:
- **2.0 g of H2**
Thus, the answer is **2.0 g of H2**.
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
From the equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
### Step 1: Calculate the moles of ammonia produced
First, we need to calculate the moles of ammonia produced (NH3). The molar mass of ammonia (NH3) is approximately:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol × 3 = 3.03 g/mol
- Total: 14.01 + 3.03 = 17.04 g/mol
Using the mass of ammonia produced (12.0 g):
\[
\text{Moles of NH}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0 \, \text{g}}{17.04 \, \text{g/mol}} \approx 0.705 \, \text{mol}
\]
### Step 2: Use the mole ratio to find moles of hydrogen
According to the equation, 2 moles of NH3 are produced from 3 moles of H2:
\[
\frac{3 \, \text{moles H}_2}{2 \, \text{moles NH}_3}
\]
Using this ratio, we can determine the moles of H2 used:
\[
\text{Moles of H}_2 = 0.705 \, \text{mol NH}_3 \times \frac{3 \, \text{mol H}_2}{2 \, \text{mol NH}_3} = 1.0575 \, \text{mol H}_2
\]
### Step 3: Calculate the mass of hydrogen used
The molar mass of hydrogen (H2) is:
- Hydrogen (H): 1.01 g/mol × 2 = 2.02 g/mol
Now we can calculate the mass of H2:
\[
\text{Mass of H}_2 = \text{moles} \times \text{molar mass} = 1.0575 \, \text{mol} \times 2.02 \, \text{g/mol} \approx 2.13 \, \text{g}
\]
Based on the options provided, this closest matches one of the responses:
- **2.0 g of H2**
Thus, the answer is **2.0 g of H2**.