One commercial system removes SO₂ emissions from smoke at 95.0°C by the following set of balanced reactions: SO₂(g) + Cl₂ $\rightarrow$ SO₂Cl₂(g)
SO₂Cl₂ + 2H₂O $\rightarrow$ H₂SO₄ + 2HCl
H₂SO₄ + Ca(OH) ₂ $\rightarrow$ 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

One commercial system removes SO₂ emissions from smoke at 95.0°C by the following set of balanced reactions: SO₂(g) + Cl₂

\rightarrow

SO₂Cl₂(g)
SO₂Cl₂ + 2H₂O

\rightarrow

H₂SO₄ + 2HCl
H₂SO₄ + Ca(OH) ₂

\rightarrow

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) ?

Quizplus · Accepted Answer

First, calculate moles of SO₂: 100 g / 64.1 g/mol = 1.56 mol. Since the reactions are 1:1, 1.56 mol of SO₂ produces 1.56 mol of CaSO₄. Convert moles of CaSO₄ to grams: 1.56 mol × 136 g/mol = 212.16 g. Considering 95% efficiency: 212.16 g × 0.95 = 201.55 g, which rounds to 202 g.

One Commercial System Removes SO2 Emissions from Smoke at 95 →\rightarrow→

One Commercial System Removes SO₂ Emissions from Smoke at 95 $\rightarrow$