how many liters of NH3, at STP, will react with 19.5 g O2 to form NO and H2O?

Answer:
10.9 L
Explanation:
Let's begin by introducing the strategy to solve this problem:
(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]