Answer :
Answer:
160.83 L, SO₂
Explanation:
From the question we are given;
Mass of PbS = 7.07 kg
Volume of Oxygen = 218 L
Temperature, T = 220°C or 493.15 K
Pressure, P = 2.00 atm
The equation for the reaction is;
2PbS + 3O₂ → 2PbO + 2SO₂
We are required to determine the volume of SO₂ formed.
We are going to use the following simple steps:
Step 1: Determine the number of moles of PbS
Number of moles = Mass ÷ Molar mass
Molar mass of PbS = 239.3 g/mol
Therefore;
Moles = 7070 g ÷ 239.3 g/mol
= 29.54 moles, PbS
Step 2: Calculate the number of moles of Oxygen
To get the number of moles of oxygen gas we are going to use ideal gas equation.
PV = nRT, R is the ideal gas constant, 0.082057 L.atm/mol.K
Therefore, moles, n = PV ÷ RT
Moles of oxygen = (2.00 atm × 218 L) ÷ (0.082057 × 493.15K)
= 10.77 moles, O₂
- Since, the number of moles of O₂ are less compared to those of PbS, then oxygen is the rate limiting reagent.
Therefore, we can determine the number of moles of SO₂.
Step 3: Determining the number of moles of SO₂
From the equation, 3 moles of oxygen reacts with PbS to produce 2 moles of SO₂.
Therefore, with 10.77 moles of Oxygen;
Moles of SO₂= Moles of O₂× (2moles SO₂/3moles O₂)
= 10.77 moles × (2/3)
= 7.18 moles, SO₂
Step 4: Determine the volume of SO₂ produced
At STP 1 mole of a gas occupies a volume of 22.4 Liters
Therefore, 7.18 moles of SO₂ will have a volume of ;
= 7.18 moles × 22.4 L/mole
= 160.832 L
= 160.83 L
Thus, 160.83 L of SO₂ are produced by the reaction.
Final answer:
The question asks for the volume of sulfur dioxide produced from roasting galena with oxygen, highlighting stoichiometric calculations in the process of lead isolation from its sulfide ore. Approximately [tex]\( 652726.70 \, \text{L} \)[/tex] of sulfur dioxide are produced at STP.
To solve this problem, we first need to write and balance the chemical equation for the reaction of galena (lead(II) sulfide, PbS) with oxygen gas to produce sulfur dioxide and lead(II) oxide (PbO):
[tex]\[ \text{PbS} + \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \][/tex]
Now, we need to determine the stoichiometry of the reaction to find the molar ratio between lead(II) sulfide and sulfur dioxide. From the balanced equation, we can see that 1 mole of lead(II) sulfide produces 1 mole of sulfur dioxide.
Next, we need to calculate the number of moles of sulfur dioxide produced from the given mass of galena (lead(II) sulfide). To do this, we'll use the molar mass of lead(II) sulfide (PbS) to convert the given mass (7.07 kg) to moles:
[tex]\[ \text{Molar mass of PbS} = \text{molar mass of Pb} + \text{molar mass of S} \][/tex]
[tex]\[ = (207.2 \, \text{g/mol}) + (32.07 \, \text{g/mol}) \][/tex]
[tex]\[ = 239.27 \, \text{g/mol} \][/tex]
Now, we convert the mass of galena to moles:
[tex]\[ \text{Moles of PbS} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ = \frac{7.07 \, \text{kg}}{239.27 \, \text{g/mol}} \][/tex]
[tex]\[ = 29535.09 \, \text{mol} \][/tex]
Since 1 mole of lead(II) sulfide produces 1 mole of sulfur dioxide, the number of moles of sulfur dioxide produced is also [tex]\( 29535.09 \) mol.[/tex]
Finally, we'll use the ideal gas law to calculate the volume of sulfur dioxide produced at STP (Standard Temperature and Pressure, [tex]\( 0^\circ \text{C} \)[/tex] and [tex]\( 1 \, \text{atm} \)):[/tex]
[tex]\[ PV = nRT \][/tex]
At STP, [tex]\( P = 1 \, \text{atm} \), \( V \)[/tex] is the volume we're solving for, [tex]\( n = 29535.09 \, \text{mol} \), \( R = 0.0821 \, \text{atm} \cdot \text{L/mol} \), and \( T = 273.15 \, \text{K} \).[/tex]
[tex]\[ V = \frac{nRT}{P} \][/tex]
[tex]\[ V = \frac{(29535.09 \, \text{mol}) \times (0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}) \times (273.15 \, \text{K})}{1 \, \text{atm}} \][/tex]
[tex]\[ V \approx 652726.70 \, \text{L} \][/tex]
Therefore, approximately [tex]\( 652726.70 \, \text{L} \)[/tex] of sulfur dioxide are produced at STP.
