Respuesta :

Answer:

10.9 L

Explanation:

Let's begin by introducing the strategy to solve this problem:

  • Write a chemical reaction and make sure it's balanced;
  • Identify STP (standard temperature and pressure, that is, T = 273.15 K and p = 1.00 atm);
  • Find moles of oxygen;
  • From the stoichiometry of the equation, identify the number of moles of ammonia;
  • Convert moles of ammonia into volume using the ideal gas law pV = nRT.

(a) The first step is already fulfilled: the reaction is balanced and it shows that 4 moles of ammonia react with 5 moles of oxygen.

(b) Given the following variables:

[tex]p=1.00 atm\\T=273.15 K\\m_O_2=19.5 g\\R=0.08206 \frac{L\cdot atm}{mol\cdot K}[/tex]

(c) Find moles of oxygen dividing mass of oxygen by its molar mass:

[tex]n_O_2=\frac{m_O_2}{M_O_2}[/tex]

Here molar mass of oxygen is:

[tex]M_O_2=32.00 g/mol[/tex]

(d) From stoichiometry, dividing moles of ammonia by its stoichiometric coefficient would be equal to the ratio of moles of oxygen divided by its stoichiometric coefficient:

[tex]\frac{n_{NH_3}}{4} =\frac{n_O_2}{5}[/tex]

Rearrange the equation to obtain moles of ammonia:

[tex]n_{NH_3}=\frac{4}{5} n_O_2[/tex]

(e) Solve pV = nRT for volume of ammonia:

[tex]pV_{NH_3}=n_{NH_3}RT\\\therefore V_{NH_3}=\frac{n_{NH_3}RT}{p}[/tex]

[tex]V_{NH_3}=\frac{\frac{4}{5}n_O_2RT }{p} =\frac{\frac{4m_O_2}{5M_O_2}RT }{p}=\frac{4m_O_2RT}{5pM_O_2}[/tex]

Substitute the given data to obtain the final answer!

[tex]{V}=\frac{4m_O_2RT}{5pM_O_2} =\frac{4\cdot19.5 g\cdot0.08206 \frac{L\cdot atm}{mol\cdot K}\cdot273.15 K }{5\cdot 1.00 atm\cdot32.00 g/mol}= 10.9 L[/tex]

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