Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g) If 5.3 L of sulfur dioxide and 4.7 L of oxygen are mixed, how much sulfur trioxide may be produced if all gases are at STP?

[tex]n_{SO_2}=\frac{5.3L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2366molSO_2\\n_{O_2}=\frac{4.7L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2098molO_2[/tex]

Now, the balanced chemical reaction turns out into:

[tex]2SO_2(g) + O_2(g) --> 2SO_3(g)[/tex]

Thus, the exact moles of oxygen that completely react with 0.2366 moles of sulfur dioxide are (limiting reagent identification):

[tex]0.2366molSO_2*\frac{1molO_2}{2molSO_2}=0.1182molO_2[/tex]

Since 0.2098 moles of oxygen are available, we stipulate the oxygen is in excess and the sulfur dioxide is the limiting reagent. In such a way, the yielded grams of sulfur trioxide turn out into:

[tex]m_{SO_3}=0.2366molSO_2*\frac{2molSO_3}{2molSO_2}*\frac{80gSO_3}{1molSO_3} \\m_{SO_3}=18.93gSO_3[/tex]

By using the ideal gas equation, one computes the volume as:

[tex]V_{SO_3}=\frac{mRT}{MP}=\frac{18.93g*0.082\frac{atm*L}{mol*K} *273.15K}{80g/mol*1atm}\\V_{SO_3}=5.3LSO_3[/tex]

It has sense for volume since the mole ratio is 2/2 between sulfur dioxide and sulfur trioxide.

Best regards.

Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: ​ SO2(g) + O2(g) → SO3(g) ​ If 5.3 L of sulfur dioxide and 4.7 L of oxygen are mixed, how much sulfur trioxide may be produced if all gases are at STP?

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Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g) If 5.3 L of sulfur dioxide and 4.7 L of oxygen are mixed, how much sulfur trioxide may be produced if all gases are at STP?