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Why is my solution for the partial pressures of two different gases incorrect?

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3

1

$begingroup$

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

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asked 17 hours ago

matryoshkamatryoshka

1184

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3

1

$begingroup$

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

share|improve this question

asked 17 hours ago

matryoshkamatryoshka

1184

New contributor

matryoshka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

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

share|improve this question

asked 17 hours ago

matryoshkamatryoshka

1184

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.

$endgroup$

share|improve this question

asked 17 hours ago

matryoshkamatryoshka

1184

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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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share|improve this question

asked 17 hours ago

matryoshkamatryoshka

1184

New contributor

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asked 17 hours ago

matryoshkamatryoshka

1184

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asked 17 hours ago

matryoshkamatryoshka

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2 Answers
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4

$begingroup$

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

share|improve this answer

answered 16 hours ago

camd92camd92

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7

$begingroup$

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).

$$P_{ce{N2O}} = frac{1}{3} times 3.70 mathrm{atm} = 1.23 mathrm{atm}$$
$$P_{ce{H2O}} = frac{2}{3} times 3.70 mathrm{atm} = 2.47 mathrm{atm}$$

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.

share|improve this answer

edited 1 hour ago

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Stian Yttervik Thanks for the edits!
  $endgroup$
  – Karsten Theis
  1 hour ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

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

share|improve this answer

answered 16 hours ago

camd92camd92

3666

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camd92 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

4

$begingroup$

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

share|improve this answer

answered 16 hours ago

camd92camd92

3666

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.

$endgroup$

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

share|improve this answer

answered 16 hours ago

camd92camd92

3666

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.

$endgroup$

share|improve this answer

answered 16 hours ago

camd92camd92

3666

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Check out our Code of Conduct.

answered 16 hours ago

camd92camd92

3666

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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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answered 16 hours ago

camd92camd92

3666

7

$begingroup$

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).

$$P_{ce{N2O}} = frac{1}{3} times 3.70 mathrm{atm} = 1.23 mathrm{atm}$$
$$P_{ce{H2O}} = frac{2}{3} times 3.70 mathrm{atm} = 2.47 mathrm{atm}$$

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.

share|improve this answer

edited 1 hour ago

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Stian Yttervik Thanks for the edits!
  $endgroup$
  – Karsten Theis
  1 hour ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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).

$$P_{ce{N2O}} = frac{1}{3} times 3.70 mathrm{atm} = 1.23 mathrm{atm}$$
$$P_{ce{H2O}} = frac{2}{3} times 3.70 mathrm{atm} = 2.47 mathrm{atm}$$

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.

share|improve this answer

edited 1 hour ago

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ Stian Yttervik Thanks for the edits!
  $endgroup$
  – Karsten Theis
  1 hour ago
  

  
  
  
  

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).

$$P_{ce{N2O}} = frac{1}{3} times 3.70 mathrm{atm} = 1.23 mathrm{atm}$$
$$P_{ce{H2O}} = frac{2}{3} times 3.70 mathrm{atm} = 2.47 mathrm{atm}$$

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.

share|improve this answer

edited 1 hour ago

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325

$endgroup$

share|improve this answer

edited 1 hour ago

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325

answered 16 hours ago

Karsten TheisKarsten Theis

2,120325