Question

Suppose 1.00 metric tons of soft coal containing 3.75 % (by mass) sulphur is burned in an electric power plant.

What volume (in L) of SO2 is produced at 60oC and 1.00 bar? L

What mass (in g) of CaO is required to "scrub" the SO2? g

What mass (in g) of CaSO3 must be dispose of? g

Answer 1

1.00 metric ton = 1000 kg

So the amount of sulphur = (3.75 x 1000) / 100 = 37.5 kg or 37500 g

S + O2 = SO2

The number of moles of S = number of moles of SO2 = 37500 / 32.065 = 1169.5 moles

Using the ideal gas law , pV = nRT

1.00 bar = 0.986923 atm

0.986923 x V = 1169.5 x 0.0821 x 333.15

V = 32411.5 L

So the volume of SO2 will be 32411.5 L

CaO + SO2 = CaSO3

equimolar amount of CaO will be required to scrub SO2.

So the number of moles of CaO = 1169.5 moles

The amount in mass (g) of CaO = 1169.5 x 56.0774 = 65582.5 g

The number of moles of CaSO3 formed will also be 1169.5

So the amount in mass = 1169.5 x 120.17 = 140538.8 grams