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Sagot :
Sure, let's work through the problem step by step.
### 1. Given Information
We have the chemical reaction:
[tex]\[ \text{NH}_4\text{NO}_3(s) \rightarrow 2 \text{N}_2(g) + 4 \text{H}_2\text{O}(g) + \text{O}_2(g) \][/tex]
We also know:
- Mass of ammonium nitrate (NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]) = 500 g
- Molar mass of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] = 80 g/mol
### 2. Calculate the Moles of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]
To find the number of moles of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex], we use the formula:
[tex]\[ \text{moles of NH}_4 \text{NO}_3 = \frac{\text{mass of NH}_4 \text{NO}_3}{\text{molar mass of NH}_4\text{NO}_3} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of NH}_4 \text{NO}_3 = \frac{500\ \text{g}}{80\ \text{g/mol}} = 6.25\ \text{moles} \][/tex]
### 3. Using Stoichiometry to Find the Moles of Gases Produced
From the balanced chemical equation, we get:
- 1 mole of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] produces 2 moles of N[tex]\(_2\)[/tex],
- 1 mole of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] produces 4 moles of H[tex]\(_2\)[/tex]O, and
- 1 mole of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] produces 1 mole of O[tex]\(_2\)[/tex].
Thus, for 6.25 moles of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]:
[tex]\[ \text{moles of N}_2 = 2 \times 6.25 = 12.5 \][/tex]
[tex]\[ \text{moles of H}_2\text{O} = 4 \times 6.25 = 25.0 \][/tex]
[tex]\[ \text{moles of O}_2 = 1 \times 6.25 = 6.25 \][/tex]
### 4. Total Moles of Gas Produced
Sum of the moles of all gases produced:
[tex]\[ \text{total moles of gas} = 12.5\ \text{moles of N}_2 + 25.0\ \text{moles of H}_2\text{O} + 6.25\ \text{moles of O}_2 \][/tex]
[tex]\[ \text{total moles of gas} = 43.75\ \text{moles} \][/tex]
### 5. Volume of Gas at STP
Standard Temperature and Pressure (STP) conditions are:
- Temperature = 273.15 K
- Pressure = 1 atm
- Ideal Gas Constant (R) = 0.0821 L·atm/(K·mol)
Using the Ideal Gas Law [tex]\( PV = nRT \)[/tex], we calculate the volume:
[tex]\[ V_{\text{STP}} = \frac{nRT}{P} \][/tex]
Plug in the values:
[tex]\[ V_{\text{STP}} = \frac{43.75\ \text{moles} \times 0.0821\ \text{L·atm/(K·mol)} \times 273.15\ \text{K}}{1\ \text{atm}} \][/tex]
[tex]\[ V_{\text{STP}} = 981.12065625\ \text{L} \][/tex]
### 6. Volume of Gas at 400 K and 150 kPa
First, convert pressure from kPa to atm (1 atm = 101.325 kPa):
[tex]\[ P = \frac{150\ \text{kPa}}{101.325\ \text{kPa/atm}} \approx 1.48\ \text{atm} \][/tex]
Using the Ideal Gas Law [tex]\( PV = nRT \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
Plug in the values:
[tex]\[ V_{\text{400K, 150kPa}} = \frac{43.75\ \text{moles} \times 0.0821\ \text{L·atm/(K·mol)} \times 400\ \text{K}}{1.48\ \text{atm}} \][/tex]
[tex]\[ V_{\text{400K, 150kPa}} \approx 970.524625\ \text{L} \][/tex]
### Final Answers
1. The total volume of gas produced at STP is approximately 981.121 liters.
2. The total volume of gas produced at 400 K and 150 kPa is approximately 970.525 liters.
### 1. Given Information
We have the chemical reaction:
[tex]\[ \text{NH}_4\text{NO}_3(s) \rightarrow 2 \text{N}_2(g) + 4 \text{H}_2\text{O}(g) + \text{O}_2(g) \][/tex]
We also know:
- Mass of ammonium nitrate (NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]) = 500 g
- Molar mass of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] = 80 g/mol
### 2. Calculate the Moles of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]
To find the number of moles of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex], we use the formula:
[tex]\[ \text{moles of NH}_4 \text{NO}_3 = \frac{\text{mass of NH}_4 \text{NO}_3}{\text{molar mass of NH}_4\text{NO}_3} \][/tex]
Substituting the given values:
[tex]\[ \text{moles of NH}_4 \text{NO}_3 = \frac{500\ \text{g}}{80\ \text{g/mol}} = 6.25\ \text{moles} \][/tex]
### 3. Using Stoichiometry to Find the Moles of Gases Produced
From the balanced chemical equation, we get:
- 1 mole of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] produces 2 moles of N[tex]\(_2\)[/tex],
- 1 mole of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] produces 4 moles of H[tex]\(_2\)[/tex]O, and
- 1 mole of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex] produces 1 mole of O[tex]\(_2\)[/tex].
Thus, for 6.25 moles of NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]:
[tex]\[ \text{moles of N}_2 = 2 \times 6.25 = 12.5 \][/tex]
[tex]\[ \text{moles of H}_2\text{O} = 4 \times 6.25 = 25.0 \][/tex]
[tex]\[ \text{moles of O}_2 = 1 \times 6.25 = 6.25 \][/tex]
### 4. Total Moles of Gas Produced
Sum of the moles of all gases produced:
[tex]\[ \text{total moles of gas} = 12.5\ \text{moles of N}_2 + 25.0\ \text{moles of H}_2\text{O} + 6.25\ \text{moles of O}_2 \][/tex]
[tex]\[ \text{total moles of gas} = 43.75\ \text{moles} \][/tex]
### 5. Volume of Gas at STP
Standard Temperature and Pressure (STP) conditions are:
- Temperature = 273.15 K
- Pressure = 1 atm
- Ideal Gas Constant (R) = 0.0821 L·atm/(K·mol)
Using the Ideal Gas Law [tex]\( PV = nRT \)[/tex], we calculate the volume:
[tex]\[ V_{\text{STP}} = \frac{nRT}{P} \][/tex]
Plug in the values:
[tex]\[ V_{\text{STP}} = \frac{43.75\ \text{moles} \times 0.0821\ \text{L·atm/(K·mol)} \times 273.15\ \text{K}}{1\ \text{atm}} \][/tex]
[tex]\[ V_{\text{STP}} = 981.12065625\ \text{L} \][/tex]
### 6. Volume of Gas at 400 K and 150 kPa
First, convert pressure from kPa to atm (1 atm = 101.325 kPa):
[tex]\[ P = \frac{150\ \text{kPa}}{101.325\ \text{kPa/atm}} \approx 1.48\ \text{atm} \][/tex]
Using the Ideal Gas Law [tex]\( PV = nRT \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
Plug in the values:
[tex]\[ V_{\text{400K, 150kPa}} = \frac{43.75\ \text{moles} \times 0.0821\ \text{L·atm/(K·mol)} \times 400\ \text{K}}{1.48\ \text{atm}} \][/tex]
[tex]\[ V_{\text{400K, 150kPa}} \approx 970.524625\ \text{L} \][/tex]
### Final Answers
1. The total volume of gas produced at STP is approximately 981.121 liters.
2. The total volume of gas produced at 400 K and 150 kPa is approximately 970.525 liters.
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