Question

Coal, which is mostly carbon, is burned in power plants in the Midwestern United States. Most coal contains some Sulfur (S₈) which is ultimately responsible for some acid rain in the eastern United States. Here's how:

1. Sulfur burns in the presence of Oxygen to create sulfur dioxide gas.

2. Sulfur dioxide gas is converted into sulfur trioxide gas when combined with oxygen in a synthesis reaction known as photooxidation (powered by the rays of the sun).

3. Sulfur trioxide reacts with water vapor in the atmosphere to create sulfuric acid which subsequently becomes aqueous in rain water and falls to the Earth.

If 0.0037% by mass of all the coal ore burned in the Midewest is actually sulfur,

(a) how many kilograms of coal ore have to burn to generate 1 kilogram of acid?

(b) how many liters of sulfur trioxide gas does this amount to (assuming STP)?

Answer 1

a) 1 kilogram of acid = 1000 grams

no of mol of acid(H2SO4) = 1000/98 = 10.2 mol

SO3 + H2O ---> H2so4

no oF mol of SO3 required = 10.2 mol

so that, no oF mol of S required = 10.2 mol

we know that, mass of S required = 10.2*32 = 326.4 g

mass of coal must burn = 326.4*100/0.0037 = 8.822*10^6 grams

                        = 8.822*10^3 kg

b) volume of SO3 = nRT/P

at STP, T = 273.15 k , P = 1 atm

        Volume(v) = 10.2*0.0821*273.15/1

                = 228.74 L