Answer:
To solve these problems, we first need to balance the chemical equation:
SF₄ + 2H₂O → SO₂ + 4HF
To find the mass of HF produced, we need to use the molar mass of SF₄ and the stoichiometry of the reaction.
Molar mass of SF₄ = 32.06 g/mol
Number of moles of SF₄ = 700 dm³ / 22.4 dm³/mol = 31.25 mol
From the balanced equation, 1 mol of SF₄ produces 4 mol of HF.
Number of moles of HF produced = 31.25 mol x 4 = 125 mol
Molar mass of HF = 20.01 g/mol
Mass of HF produced = 125 mol x 20.01 g/mol = 2502.5 g or 2.5025 kg
To find the mass of SO₂ produced, we can use the same approach.
From the balanced equation, 1 mol of SF₄ produces 1 mol of SO₂.
Number of moles of SO₂ produced = 31.25 mol x 1 = 31.25 mol
Molar mass of SO₂ = 64.06 g/mol
Mass of SO₂ produced = 31.25 mol x 64.06 g/mol = 2001.56 g or 2.00156 kg
To find the volume of SO₂ released, we need to use the ideal gas law.
PV = nRT
Assuming the reaction takes place at standard temperature and pressure (STP), we have:
P = 1 atm
V = ?
n = 31.25 mol
R = 0.08206 L·atm/K·mol
T = 273 K
Solving for V, we get:
V = nRT/P = 31.25 mol x 0.08206 L·atm/K·mol x 273 K / 1 atm = 684.99 L or 0.685 m³
To find the mass of H₂O needed, we can use the stoichiometry of the reaction.
From the balanced equation, 1 mol of SF₄ reacts with 2 mol of H₂O.
Number of moles of H₂O needed = 31.25 mol x 2 = 62.5 mol
Molar mass of H₂O = 18.02 g/mol
Mass of H₂O needed = 62.5 mol x 18.02 g/mol = 1126.25 g or 1.12625 kg
The percent yield is calculated as the actual yield divided by the theoretical yield, multiplied by 100%.
Theoretical yield of HF = 125 mol x 20.01 g/mol = 2502.5 g or 2.5025 kg (same as in part 1)
Actual yield of HF = 1500 g
Percent yield = (actual yield / theoretical yield) x 100% = (1500 g / 2502.5 g) x 100% = 59.96%
Therefore, the percent yield of the reaction is approximately 59.96%.