One commercial system removes SO₂ emissions from smoke at 95.0°C by the following set of balanced reactions:
SO₂(g) + Cl₂ → SO₂Cl₂(g)
SO₂Cl₂ + 2H₂O → H₂SO₄ + 2HCl
H₂SO₄ + Ca(OH) ₂ → CaSO₄(s) + 2H₂O
Assuming the process is 95.0% efficient, how many grams of CaSO₄ may be produced from 1.00 × 10² g of SO₂ (molar masses: SO₂, 64.1 g/mol; CaSO₄, 136 g/mol) ?

Question

Quizplus · Accepted Answer

First, we need to find the moles of SO₂ in 1.00 × 10² g: moles of SO₂ = (1.00 × 10² g) / (64.1 g/mol) = 1.56 × 10² mol Next, we need to use stoichiometry to find the moles of CaSO₄ that can be produced. From the balanced equations, we see that 1 mol of SO₂ can produce 1 mol of CaSO₄: SO₂ + Cl₂ → SO₂Cl₂ SO₂Cl₂ + 2H₂O → H₂SO₄ + 2HCl H₂SO₄ + Ca(OH)₂ → CaSO₄(s) + 2H₂O So, the moles of CaSO₄ that can be produced are also 1.56 × 10² mol. However, the process is only 95.0% efficient, meaning that we will not get a 100% yield of CaSO₄. We can calculate the actual yield by multiplying the theoretical yield (1.56 × 10² mol) by the efficiency: actual yield = (1.56 × 10² mol) x 0.95 = 1.48 × 10² mol Finally, we can convert the moles of CaSO₄ to grams: mass of CaSO₄ = (1.48 × 10² mol) x (136 g/mol) = 2.01 × 10³ g Therefore, the answer is 202 g.

One Commercial System Removes SO2 Emissions from Smoke at 95

One Commercial System Removes SO₂ Emissions from Smoke at 95