<h3>
Answer:</h3>
#a. Theoretical yield = 31.6 g
#b. Actual yield = 25.72 g
<h3>
Explanation:</h3>
The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;
SO₂(g) + H₂O(l) → H₂SO₃(aq)
The percent yield of H₂SO₃ is 81.4%
Mass of water that reacted is 6.94 g
#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps
Step 1: Calculate the moles of water
Molar mass of water = 18.02 g/mol
Mass of water = 6.94 g
But, moles = Mass/molar mass
Moles of water = 6.94 g ÷ 18.02 g/mol
= 0.385 mol
Step 2: Calculate moles of H₂SO₃
From the equation, the mole ratio of water to H₂SO₃ is 1 : 1
Therefore, moles of water = moles of H₂SO₃
Hence, moles of H₂SO₃ = 0.385 mol
Step 3: Theoretical mass of H₂SO₃
Mass = moles × Molar mass
Molar mass of H₂SO₃ = 82.08 g/mol
Number of moles of H₂SO₃ = 0.385 mol
Therefore;
Theoretical mass of H₂SO₃ = 0.385 mol × 82.08 g/mol
= 31.60 g
Thus, the theoretical yield of H₂SO₃ is 31.6 g
<h3>#b. Calculating the actual yield</h3>
We need to calculate the actual yield
Percent yield of H₂SO₃ is 81.4%
Theoretical yield is 31.60 g
But; Percent yield = (Actual yield/theoretical yield)×100
Therefore;
Actual yield = Percent yield × theoretical yield)÷ 100
= (81.4 % × 31.6) ÷ 100
= 25.72 g
The percent yield of H₂SO₃ is 25.72 g
