Use stoicheometry

Question

A certain sample of coal contains 1.4% sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxidr is treated with calcium oxide (CaO) to form calcium sulfite (CaSO3). Calculate the daily mass (In kilograms) of CaO needed if a power plant uses 9.10 x 10^6 kg of coal per day.

Answer 1

Lets write out chemical reaction equations first

1) S(s) + O₂(g) → SO₂(g)

2) SO₂(g) + CaO(s) → CaSO₃ (s)

9.10 x 10^6 kg of coal per day so 1.4% of this is 9.10 x 10⁶(0.014) = 127,400kg of Sulfur is burned each day

Convert 127,400 kg of Sulfur into grams then into moles

127,400kg * (1000g/1kg) * (1mol Sulfur/ 32.06g Sulfur) = 3.97 * 10⁶ mol Sulfur

2) SO₂(g) + CaO(s) → CaSO₃ (s)

Stoichiometric mole ratio between SO₂ (g) and CaO(s) is 1:1 therefore 1 mole of CaO(s) reacts with 1 mole of SO₂ (g) to produce 1 mole of CaSO₃ (s).

3.97 * 10⁶ mol Sulfur * (1 mol CaO / 1 mol Sulfur) = 3.97 * 10⁶ mol CaO

3.97 * 10⁶ mol CaO * ( 56.08g CaO / 1mol CaO) = 2.23 * 10⁸g = 2.23 * 10⁵kg of CaO needed