Respuesta :
Answer : The volume of [tex]SO_2[/tex] produced are 90.7 liters.
Explanation :
The balanced chemical reaction will be:
[tex]2PbS+3O_2\rightarrow 2PbO+2SO_2[/tex]
First we have to calculate the moles of PbS.
[tex]\text{Moles of }PbS=\frac{\text{Mass of }PbS}{\text{Molar mass of }PbS}[/tex]
Molar mass of PbS = 239.26 g/mole
[tex]\text{Moles of }PbS=\frac{5.05kg}{239.26}=\frac{5.05\times 1000g}{239.26}=21.1mole[/tex]
Now we have to calculate the moles of [tex]O_2[/tex] by using ideal gas equation.
Using ideal gas equation :
[tex]PV=nRT[/tex]
where,
P = Pressure of [tex]O_2[/tex] gas = 2.00 atm
V = Volume of [tex]O_2[/tex] gas = 123 L
n = number of moles [tex]O_2[/tex] = ?
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]O_2[/tex] gas = [tex]220^oC=273+220=493K[/tex]
Putting values in above equation, we get:
[tex]2.00atm\times 123L=n\times (0.0821L.atm/mol.K)\times 493K[/tex]
[tex]n=6.07mole[/tex]
The number of moles of [tex]O_2[/tex] is, 6.07 mole
Now we have to calculate the limiting and excess reagent.
From the balanced reaction we conclude that
As, 3 mole of [tex]O_2[/tex] react with 2 mole of [tex]PbS[/tex]
So, 6.07 moles of [tex]O_2[/tex] react with [tex]\frac{6.07}{3}\times 2=4.05[/tex] moles of [tex]PbS[/tex]
From this we conclude that, [tex]PbS[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]SO_2[/tex]
From the reaction, we conclude that
As, 3 mole of [tex]O_2[/tex] react to give 2 mole of [tex]SO_2[/tex]
So, 6.07 moles of [tex]O_2[/tex] react to give [tex]\frac{6.07}{3}\times 2=4.05[/tex] moles of [tex]SO_2[/tex]
Now we have to calculate the volume of [tex]SO_2[/tex] produced at STP.
As, 1 mole of [tex]SO_2[/tex] contains 22.4 L volume of [tex]SO_2[/tex]
So, 4.05 mole of [tex]SO_2[/tex] contains [tex]4.05\times 22.4=90.7L[/tex] volume of [tex]SO_2[/tex]
Therefore, the volume of [tex]SO_2[/tex] produced are 90.7 liters.
