Answer:
The idea here is that if all the gases that take part in the reaction are kept ... the reaction must consume 3 moles of hydrogen gas and 1 mole of nitrogen gas.
Explanation:
160.808 L of ammonia is formed when (5000.0 L) of hydrogen gas at 100.0 K and 20.0 atm is present.
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given:
V=5000.0 L
T=100.0 K
P=20.0 atm
[tex]R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}[/tex]
Putting value in the given equation:
[tex]\frac{PV}{RT}=n_{H_2}[/tex]
[tex]\frac{20.0 atm X 5000.0 L}{ 0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 100.0 K}=n_{H_2}[/tex]
[tex]n_{H_2}=12180.268 mole[/tex]
Now,
[tex]N_2 +3H_2[/tex]→ [tex]2NH_3[/tex]
[tex]N_2[/tex] is excess
[tex]\frac{n_{H_2}}{3} =\frac{n_{NH_3}}{2}[/tex]
[tex]n_{NH_3} = \frac{2 X 12180.268 mole}{3}[/tex]
[tex]n_{NH_3} = 8120.179 mole[/tex]
Molar mass of [tex]NH_3[/tex] =17.031 g/mole
[tex]Mass_{NH_3}[/tex]=8120.179 x 17.031 = 138294.76 g
[tex]Density_{NH_3}[/tex] = [tex]mass_{NH_3}[/tex] ÷ [tex]Volume_{NH_3}[/tex]
[tex]Volume_{NH_3}[/tex] = 138294.76 g ÷ 0.860 g/mL
[tex]Volume_{NH_3}[/tex] = 160.808 L
Hence, 160.808 L of ammonia is formed when (5000.0 L) of hydrogen gas at 100.0 K and 20.0 atm is present.
Learn more about the ideal gas here:
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