Answer :
Answer : The volume of sulfur dioxide measured at STP is, 113.6 L
Explanation :
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 = 239.27 g/mol
[tex]\text{Moles of }PbS=\frac{6.61kg}{239.27g/mol}=\frac{6610g}{239.27g/mol}=27.62mol[/tex]
Now we have to calculate the moles of oxygen gas.
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 = 154 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 154L=n\times (0.0821L.atm/mol.K)\times 493K[/tex]
[tex]n=7.61mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be:
[tex]2PbS(s)+3O_2(g)\rightarrow 2PbO(s)+2SO_2(g)[/tex]
From the balanced chemical reaction we conclude that,
As, 3 moles of [tex]O_2[/tex] react with 2 moles of PbS
So, 7.61 moles of [tex]O_2[/tex] react with [tex]\frac{2}{3}\times 7.61=5.07[/tex] moles of PbS
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, 7.61 mole of [tex]O_2[/tex] react to give [tex]\frac{2}{3}\times 7.61=5.07[/tex] mole of [tex]SO_2[/tex]
Now we have to calculate the volume of [tex]SO_2[/tex] at STP.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of [tex]SO_2[/tex] gas = 1.00 atm
V = Volume of [tex]SO_2[/tex] gas = ?
n = number of moles [tex]SO_2[/tex] = 5.07 mol
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]SO_2[/tex] gas = 273 K
Putting values in above equation, we get:
[tex]1.00atm\times V=5.07mol\times (0.0821L.atm/mol.K)\times 273K[/tex]
[tex]V=113.6L[/tex]
Thus, the volume of sulfur dioxide measured at STP is, 113.6 L
The volume of the ideal gas can be given by moles, gas constant, temperature and pressure. The volume of sulfur dioxide at STP is, 113.6 L.
What is STP?
STP is the standard temperature and the pressure that can be used for the comparative estimation of the gases. At STP the temperature is 273 K and pressure is 1 atm.
Moles of PbS is calculated as:
[tex]\begin{aligned}\rm Moles &= \rm \dfrac{Mass}{Molar\;mass}\\\\&= \dfrac{6610}{239.27}\\\\&= 27.62\;\rm mol\end{aligned}[/tex]
Moles of oxygen gas can be calculated by the ideal gas equation as:
The pressure of oxygen gas (P) = 2.00 atm
The volume of oxygen (V) = 154 L
Gas constant (R) = 0.0821
Temperature of the gas(T) = 493 K
Substituting values in the ideal gas equation n can be calculated as:
[tex]\rm n &= \rm \dfrac{PV}{RT}\\\\\\&= \dfrac{2\times 154}{0.0821\times 493}\\\\\\&= 7.61\;\rm mol\end{aligned}[/tex]
The balanced reaction can be shown as:
[tex]\rm 2PbS + 3 O_{2} \rightarrow 2PbO + 2SO_{2}[/tex]
From this, it can be said that 3 moles of oxygen react with 2 moles of PbS so, 7.61 moles of oxygen will react with 5.07 moles of PbS.
The moles of the reaction depicts that PbS is the excess reagent and oxygen is the limiting reagent.
Moles of Sulfur dioxide is calculated as:
3 moles of oxygen = 2 moles of Sulfur dioxide
So, 7.61 moles of oxygen = 5.07 moles of Sulfur dioxide
Using the ideal gas equation the volume of Sulfur dioxide can be calculated as:
[tex]\begin{aligned}\rm V&= \rm \dfrac{nRT}{P}\\\\&= \dfrac{5.07\times 0.0821\times 273}{1}\\\\&= 113.63\;\rm L\end{aligned}[/tex]
Therefore, the volume of Sulfur dioxide at STP will be 113.6 L.
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