Final answer:
To find the volume of SO₂ produced by burning 1.00 kg of sulfur in excess oxygen at 343 °C and 1.21 atm, we can use the ideal gas law equation. First, find the number of moles of SO₂ produced. Then, plug in the values into the ideal gas law equation to find the volume of SO₂.
Step-by-step explanation:
To find the volume of SO₂ produced by burning 1.00 kg of sulfur in excess oxygen at 343 °C and 1.21 atm, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for V: V = (nRT) / P.
First, we need to find the number of moles of SO₂ produced. To do this, we can use the molar mass of sulfur and the given mass of sulfur. The molar mass of sulfur is 32.07 g/mol. The number of moles of sulfur is therefore 1.00 kg / 32.07 g/mol = X moles of S.
Since SO₂ is 2 moles of oxygen for every mole of sulfur, the number of moles of SO₂ is X moles of S x 2 moles of O / 1 mole of S = X moles of SO₂. Now we can plug in the values into the ideal gas law equation to find the volume of SO₂:
V = (X moles of SO₂ x 0.0821 L x atm / mol x K x 343 K) / 1.21 atm.
This gives us the volume of SO₂ in liters.