Question

Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mmHg), what is the limiting reactant and the theoretical yield of SO3? If 187.2 mL of SO3 is collected (measured at 315 K and 50.0 mmHg), what is the percent yield for the reaction?

Answer 1

Answer:

a. Oxygen is the limiting reagent. $n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3$

b. $Y=58.9$%

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

$n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2$

Afterwards, by considering the properly balanced chemical reaction:

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

We compute the oxygen's moles that completely reacts with the previously computed $7.273x10^{-3}$ moles of $SO_2$ as follows:

$7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2$

That result let us know that the oxygen is the limiting reagent since just $4.048x10^{-4}$ moles are available in comparison with the $3.6365x10^{-3}$ moles that completely would react with $7.273x10^{-3}$ moles of $SO_2$.

Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:

$n_{SO_3}^{Theoretical}=4.048x10^{-4}molO_2*\frac{2molSO_3}{1molO_2} =8.096x10^{-4}mol SO_3$

b. Percent yield.

At first, we must compute the collected (real) moles of sulfur trioxide:

$n_{SO_3}^{real}=\frac{PV}{RT}=\frac{0.0658atm*0.1872L}{0.082\frac{atm*L}{mol*K}*315K} =4.769x10^{-4}molSO_3$

Finally, we compute the percent yield:

$Y=\frac{n_{SO_3}^{real}}{n_{SO_3}^{Theoretical}} *100$%

$Y=\frac{4.769x10^{-4}mol SO_3}{8.096x10^{-4}mol SO_3} *100$%

$Y=58.9$%

Best regards.

Answer 2

Answer:The limiting reactants is SO2

The theoretical yield of SO3 is 285.5

The %yield of SO3 is 65.6%

Explanation:from Dalton's atomic principle ,when gases combines in a reaction ,it is always in a simple ratio of their Volumes.

From the balnaced equation 2SO2 +O2___2SO3,we can infer that 2volume of SO2 requires 1Volume of O2,so if O2 present is 142.75ml it will need 142.75×2 Vol of SO2 to react and produce 2 vol of SO3.

Therefore,we can infer that O2 is surplus in the reaction while SO2 is in short supply.

The theoretical yield from Dalton's principle will be 2× 142.75=285.5ml.

The percentage yield of SO3 will be the real yield/expected yield ×100=187.2/285.5×100=65.6%