1.Ammonia, NH3(g), reacts with oxygen, O2 (g), to form nitrogen dioxide, NO2(g), and water, H2O(g), according to the following equation. 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) You react ammonia, and oxygen,and at the end of the experiment you find that you produced 27.0 g of water, and have some ammonia left over. Determine the mass of oxygen reacted. Please show all of your work for credit.
2.Reacting 3.00 mol nitrogen gas with 6.00 mol of hydrogen gas will produce how many moles of ammonia according to the following balanced chemical equation?
N2(g) + 3H2(g) ® 2NH3(g)
The Correct Answer and Explanation is :
Here’s the detailed solution to both problems:
1. Mass of oxygen reacted
The balanced equation is:
[ 4\text{NH}_3(g) + 7\text{O}_2(g) \rightarrow 4\text{NO}_2(g) + 6\text{H}_2\text{O}(g) ]
We are given:
- Mass of (\text{H}_2\text{O} = 27.0 \, \text{g})
- Ammonia is in excess, so oxygen is the limiting reagent.
Step 1: Convert mass of (\text{H}_2\text{O}) to moles
The molar mass of (\text{H}_2\text{O}) is:
[
\text{H}_2\text{O} = 2(1.008) + 16.00 = 18.016 \, \text{g/mol}
]
[
\text{Moles of H}_2\text{O} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{27.0 \, \text{g}}{18.016 \, \text{g/mol}} = 1.499 \, \text{mol}
]
Step 2: Use stoichiometry to find moles of (\text{O}_2) reacted
From the balanced equation, the molar ratio between (\text{H}_2\text{O}) and (\text{O}_2) is (6:7). Thus:
[
\text{Moles of O}_2 = \frac{\text{Moles of H}_2\text{O} \times 7}{6} = \frac{1.499 \times 7}{6} = 1.749 \, \text{mol}
]
Step 3: Convert moles of (\text{O}_2) to mass
The molar mass of (\text{O}_2) is:
[
\text{O}_2 = 2(16.00) = 32.00 \, \text{g/mol}
]
[
\text{Mass of O}_2 = \text{Moles of O}_2 \times \text{Molar Mass of O}_2 = 1.749 \, \text{mol} \times 32.00 \, \text{g/mol} = 55.97 \, \text{g}
]
Final Answer for Part 1:
The mass of oxygen reacted is:
[
\boxed{55.97 \, \text{g}}
]
2. Moles of ammonia produced
The balanced equation is:
[
\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)
]
We are given:
- (\text{N}_2 = 3.00 \, \text{mol})
- (\text{H}_2 = 6.00 \, \text{mol})
Step 1: Determine the limiting reagent
From the balanced equation, 1 mol of (\text{N}_2) reacts with 3 mol of (\text{H}_2). The required (\text{H}_2) for 3.00 mol of (\text{N}_2) is:
[
\text{Required H}_2 = 3.00 \, \text{mol} \times 3 = 9.00 \, \text{mol}
]
Since only 6.00 mol of (\text{H}_2) is available, (\text{H}_2) is the limiting reagent.
Step 2: Calculate moles of (\text{NH}_3) produced
From the balanced equation, 3 mol of (\text{H}_2) produces 2 mol of (\text{NH}_3). Thus:
[
\text{Moles of NH}_3 = \frac{\text{Moles of H}_2 \times 2}{3} = \frac{6.00 \times 2}{3} = 4.00 \, \text{mol}
]
Final Answer for Part 2:
The moles of ammonia produced is:
[
\boxed{4.00 \, \text{mol}}
]
Explanation for Part 2 (300 words)
The reaction between nitrogen gas ((\text{N}_2)) and hydrogen gas ((\text{H}_2)) follows the balanced equation:
[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
]
This equation shows that 1 mol of nitrogen reacts with 3 mol of hydrogen to produce 2 mol of ammonia. In stoichiometric calculations, the limiting reagent determines the maximum amount of product formed. Here, we begin with 3.00 mol of (\text{N}_2) and 6.00 mol of (\text{H}_2). To check which reactant limits the reaction, we calculate the amount of (\text{H}_2) required to react fully with (\text{N}_2). For 3.00 mol of (\text{N}_2), the reaction requires (3.00 \times 3 = 9.00 \, \text{mol}) of (\text{H}_2). However, only 6.00 mol of (\text{H}_2) is available, making (\text{H}_2) the limiting reagent. Using the stoichiometric relationship between (\text{H}_2) and (\text{NH}_3) (3:2), we find that 6.00 mol of (\text{H}_2) produces:
[
\frac{6.00 \, \text{mol} \times 2}{3} = 4.00 \, \text{mol of NH}_3
]
Thus, the maximum yield of ammonia is 4.00 mol. This calculation highlights the importance of identifying the limiting reagent in reactions, as it dictates the product yield, regardless of the excess amount of other reactants.