<h3>
Answer:</h3>
160.83 L, SO₂
<h3>
Explanation:</h3>
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:
<h3>Step 1: Determine the number of moles of PbS </h3>
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
<h3>Step 2: Calculate the number of moles of Oxygen </h3>
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.
<h3>Step 3: Determining the number of moles of SO₂</h3>
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
<h3>Step 4: Determine the volume of SO₂ produced </h3>
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.
