Respuesta :
Answer: The volume of sulfur dioxide measured at STP is 125.44 L
Explanation:
- For galena:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of galena = 3.73 kg = 3730 g (Conversion factor: 1 kg = 1000 g)
Molar mass of galena = 239.3 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of galena}=\frac{3730g}{239.3g/mol}=15.59mol[/tex]
- For oxygen gas:
To calculate the amount of oxygen gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 2.00 atm
V = Volume of the gas = 170 L
T = Temperature of the gas = [tex]220^oC=[220+273]K=493K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of oxygen gas = ?
Putting values in above equation, we get:
[tex]2.00atm\times 170L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 493\\\\n=\frac{2.00\times 170}{0.0821\times 493}=8.4mol[/tex]
The chemical equation for the reaction of lead (II) sulfide and oxygen gas follows:
[tex]2PbS+3O_2\rightarrow 2PbO+2SO_2[/tex]
By Stoichiometry of the reaction:
3 moles of oxygen gas reacts with 2 moles of galena
So, 8.4 moles of oxygen gas will react with = [tex]\frac{2}{3}\times 8.4=5.6mol[/tex] of galena
As, given amount of lead (II) sulfide is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 moles of oxygen gas produces 2 moles of sulfur dioxide
So, 8.4 moles of oxygen gas will produce = [tex]\frac{2}{3}\times 8.4=5.6moles[/tex] of sulfur dioxide
At STP:
1 mole of a gas occupies 22.4 L of volume
So, 5.6 moles of sulfur dioxide will occupy = [tex]\frac{22.4L}{1mol}\times 5.6mol=125.44L[/tex] of volume
Hence, the volume of sulfur dioxide measured at STP is 125.44 L
