In the contact process, sulfuric acid is manufactured by first oxidizing SO2 to SO3, which is then reacted with water. The reaction of SO2 with O2 is
2SO2(g) 1 O2(g) m 2SO3(g)
A 2.000-L flask was filled with 0.0400 mol SO2 and 0.0200 mol O2. At equilibrium at 900 K, the flask contained 0.0296 mol SO3. How many moles of each substance were in the flask at equilibrium?

The moles of each substance at equilibrium has been [tex]\rm \bold{SO_2}[/tex] has been 0.01216 mol, [tex]\rm O_2[/tex] has been 0.01568 mol, and [tex]\rm SO_3[/tex] has been 0.02792 mol.

The contact process has been the industrial process of the synthesis of sulfuric acid. The balanced equation for the reaction has been:

[tex]\rm 2\;SO_2\;+\;O_2\;\rightarrow\;2\;SO_3[/tex]

The ICE table for the reaction has been given as:

[tex]\rm SO_2[/tex] [tex]\rm O_2[/tex] [tex]\rm SO_3[/tex]

I 0.04 0.0296 0

C -2x -x 2x

E 0.04 - 2x 0.0296 - x 2x

At the equilibrium, number of moles of reactants and product has been equal. Thus, for the reaction:

[tex]\rm SO_2\;+\;O_2\;=\;SO_3[/tex]

[tex]0.04 - 2x + 0.0296 - x = 2x \\0.04 + 0.0296 = 2x + x + 2x\\0.696 = 5x\\x = 0.01392[/tex]

The moles of each substance at equilibrium has been:

Moles of [tex]\rm SO_2[/tex] has been:

[tex]\rm SO_2\;=\;0.04\;-\;2x\\\\SO_2\;=\;0.004\;-\;2(0.1392)\\\\SO_2\;=\;0.01216\;mol[/tex]

Moles of [tex]\rm O_2[/tex] has been:

[tex]\rm O_2\;=\;0.0296\;-\;x\\\\O_2\;=\;0.0296\;-\;0.1392\\\\O_2\;=\;0.01568\;mol[/tex]

Moles of [tex]\rm SO_3[/tex] has been:

[tex]\rm SO_3\;=\;2x\\\\SO_2\;=\;2(0.1392)\\\\SO_2\;=\;0.02792\;mol[/tex]

The moles of each substance at equilibrium has been [tex]\rm \bold{SO_2}[/tex] has been 0.01216 mol, [tex]\rm O_2[/tex] has been 0.01568 mol, and [tex]\rm SO_3[/tex] has been 0.02792 mol.

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https://brainly.com/question/921246