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












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.






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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







    share|improve this question







    New contributor




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







      share|improve this question







      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






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      asked 2 hours ago









      matryoshkamatryoshka

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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






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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.






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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






              share|improve this answer








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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.






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                2












                $begingroup$

                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








                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.






                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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








                  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







                  share|improve this answer








                  New contributor




                  camd92 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  answered 1 hour ago









                  camd92camd92

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                      1












                      $begingroup$

                      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.






                      share|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        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.






                        share|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          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.






                          share|improve this answer









                          $endgroup$



                          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.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 1 hour ago









                          Karsten TheisKarsten Theis

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