Question

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?

Answer 1

`m_(SO_3)=18.93gSO_3`

`V_(SO_3)=5.3LSO_3`

Step-by-step explanation:

Hello,

STP conditions are P=1 atm and T=273.15 K, thus, the reacting moles are:

`n_(SO_2)=(5.3L*1atm)/(0.082(atm*L)/(mol*K)*273.15K)=0.2366molSO_2\\n_(O_2)=(4.7L*1atm)/(0.082(atm*L)/(mol*K)*273.15K)=0.2098molO_2`

Now, the balanced chemical reaction turns out into:

`2SO_2(g) + O_2(g) --> 2SO_3(g)`

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

`0.2366molSO_2*(1molO_2)/(2molSO_2)=0.1182molO_2`

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:

`m_(SO_3)=0.2366molSO_2*(2molSO_3)/(2molSO_2)*(80gSO_3)/(1molSO_3) \\m_(SO_3)=18.93gSO_3`

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

`V_(SO_3)=(mRT)/(MP)=(18.93g*0.082(atm*L)/(mol*K) *273.15K)/(80g/mol*1atm)\\V_(SO_3)=5.3LSO_3`

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

Best regards.