Why is my solution for the partial pressures of two different gases incorrect?Why does the cathode ray tube...
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Why is my solution for the partial pressures of two different gases incorrect?
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$begingroup$
I have the chemical equation: NH4NO3(s) $rightarrow$ 2H2O (g) + N2O (g). There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00$ C until the total pressure is $3.70$ atm.
This is what I tried:
Because there are 2 more H2O than N2O I said that there are $9.06g$ of H2O and $3.02g$ of N2O.
$$9.06gtimes {molover 18.016g}=0.503 mol H_2O$$
$$3.062times {molover 44.02g}=0.0686 mol N_2O$$
I then found the mole ratio of both compounds:
$$X_{H_{2}O}={0.503over 0.572}=0.879 $$
$$X_{N_{2}O}={0.0686over 0.572}=0.1199$$
Then I used the relationship between mole ratios and partial pressure $X_i={P_i over P_{tot}}$, where $P_{tot}=3.70atm $:
$$P_{H_2O}=(0.879)(3.70atm)=3.25atm$$
$$P_{N_2O}=(0.1199)(3.70atm)=0.444atm$$
According to the answer solution for this problem $P_{H_2O}=2.47atm$ and $P_{N_2O}=1.23atm$. What am I doing wrong? I can't figure out how to get the correct solution.
pressure
asked 4 hours ago
matryoshkamatryoshka
1084
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$begingroup$
I have the chemical equation: NH4NO3(s) $rightarrow$ 2H2O (g) + N2O (g). There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00$ C until the total pressure is $3.70$ atm.
This is what I tried:
Because there are 2 more H2O than N2O I said that there are $9.06g$ of H2O and $3.02g$ of N2O.
$$9.06gtimes {molover 18.016g}=0.503 mol H_2O$$
$$3.062times {molover 44.02g}=0.0686 mol N_2O$$
I then found the mole ratio of both compounds:
$$X_{H_{2}O}={0.503over 0.572}=0.879 $$
$$X_{N_{2}O}={0.0686over 0.572}=0.1199$$
Then I used the relationship between mole ratios and partial pressure $X_i={P_i over P_{tot}}$, where $P_{tot}=3.70atm $:
$$P_{H_2O}=(0.879)(3.70atm)=3.25atm$$
$$P_{N_2O}=(0.1199)(3.70atm)=0.444atm$$
According to the answer solution for this problem $P_{H_2O}=2.47atm$ and $P_{N_2O}=1.23atm$. What am I doing wrong? I can't figure out how to get the correct solution.
pressure
asked 4 hours ago
matryoshkamatryoshka
1084
New contributor
matryoshka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I have the chemical equation: NH4NO3(s) $rightarrow$ 2H2O (g) + N2O (g). There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00$ C until the total pressure is $3.70$ atm.
This is what I tried:
Because there are 2 more H2O than N2O I said that there are $9.06g$ of H2O and $3.02g$ of N2O.
$$9.06gtimes {molover 18.016g}=0.503 mol H_2O$$
$$3.062times {molover 44.02g}=0.0686 mol N_2O$$
I then found the mole ratio of both compounds:
$$X_{H_{2}O}={0.503over 0.572}=0.879 $$
$$X_{N_{2}O}={0.0686over 0.572}=0.1199$$
Then I used the relationship between mole ratios and partial pressure $X_i={P_i over P_{tot}}$, where $P_{tot}=3.70atm $:
$$P_{H_2O}=(0.879)(3.70atm)=3.25atm$$
$$P_{N_2O}=(0.1199)(3.70atm)=0.444atm$$
According to the answer solution for this problem $P_{H_2O}=2.47atm$ and $P_{N_2O}=1.23atm$. What am I doing wrong? I can't figure out how to get the correct solution.
pressure
asked 4 hours ago
matryoshkamatryoshka
1084
New contributor
matryoshka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have the chemical equation: NH4NO3(s) $rightarrow$ 2H2O (g) + N2O (g). There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00$ C until the total pressure is $3.70$ atm.
This is what I tried:
Because there are 2 more H2O than N2O I said that there are $9.06g$ of H2O and $3.02g$ of N2O.
$$9.06gtimes {molover 18.016g}=0.503 mol H_2O$$
$$3.062times {molover 44.02g}=0.0686 mol N_2O$$
I then found the mole ratio of both compounds:
$$X_{H_{2}O}={0.503over 0.572}=0.879 $$
$$X_{N_{2}O}={0.0686over 0.572}=0.1199$$
Then I used the relationship between mole ratios and partial pressure $X_i={P_i over P_{tot}}$, where $P_{tot}=3.70atm $:
$$P_{H_2O}=(0.879)(3.70atm)=3.25atm$$
$$P_{N_2O}=(0.1199)(3.70atm)=0.444atm$$
According to the answer solution for this problem $P_{H_2O}=2.47atm$ and $P_{N_2O}=1.23atm$. What am I doing wrong? I can't figure out how to get the correct solution.
pressure
pressure
asked 4 hours ago
matryoshkamatryoshka
1084
New contributor
matryoshka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
matryoshkamatryoshka
1084
New contributor
matryoshka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
matryoshkamatryoshka
1084
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matryoshka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
matryoshkamatryoshka
1084
asked 4 hours ago
matryoshkamatryoshka
1084
1084
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2 Answers
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$begingroup$
You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $ce{NH4NO3}$, then you have:
$$ce{12.8 g NH4NO3}timesfrac{ce{1 mol NH4NO3}}{ce{80.04 g NH4NO3}}=ce{0.160 mol NH4NO3}$$
Since $ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $ce{H2O}$ and $ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer
answered 3 hours ago
camd92camd92
1316
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The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).
The bottom line is that chemical equations (i.e. stoichiometry) describes the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
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2 Answers
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2 Answers
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$begingroup$
You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $ce{NH4NO3}$, then you have:
$$ce{12.8 g NH4NO3}timesfrac{ce{1 mol NH4NO3}}{ce{80.04 g NH4NO3}}=ce{0.160 mol NH4NO3}$$
Since $ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $ce{H2O}$ and $ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer
answered 3 hours ago
camd92camd92
1316
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camd92 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $ce{NH4NO3}$, then you have:
$$ce{12.8 g NH4NO3}timesfrac{ce{1 mol NH4NO3}}{ce{80.04 g NH4NO3}}=ce{0.160 mol NH4NO3}$$
Since $ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $ce{H2O}$ and $ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer
answered 3 hours ago
camd92camd92
1316
New contributor
camd92 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $ce{NH4NO3}$, then you have:
$$ce{12.8 g NH4NO3}timesfrac{ce{1 mol NH4NO3}}{ce{80.04 g NH4NO3}}=ce{0.160 mol NH4NO3}$$
Since $ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $ce{H2O}$ and $ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer
answered 3 hours ago
camd92camd92
1316
New contributor
camd92 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $ce{NH4NO3}$, then you have:
$$ce{12.8 g NH4NO3}timesfrac{ce{1 mol NH4NO3}}{ce{80.04 g NH4NO3}}=ce{0.160 mol NH4NO3}$$
Since $ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $ce{H2O}$ and $ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer
answered 3 hours ago
camd92camd92
1316
New contributor
camd92 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 3 hours ago
camd92camd92
1316
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answered 3 hours ago
camd92camd92
1316
answered 3 hours ago
camd92camd92
1316
1316
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$begingroup$
The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).
The bottom line is that chemical equations (i.e. stoichiometry) describes the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
$endgroup$
add a comment |
$begingroup$
The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).
The bottom line is that chemical equations (i.e. stoichiometry) describes the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
$endgroup$
add a comment |
$begingroup$
The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).
The bottom line is that chemical equations (i.e. stoichiometry) describes the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
$endgroup$
The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).
The bottom line is that chemical equations (i.e. stoichiometry) describes the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
answered 2 hours ago
Karsten TheisKarsten Theis
2,070325
2,070325
add a comment |
add a comment |
matryoshka is a new contributor. Be nice, and check out our Code of Conduct.
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matryoshka is a new contributor. Be nice, and check out our Code of Conduct.
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