Answer to Question #260314 in General Chemistry for Blabla

Question

As a young DENR lab analyst, you know that sulfur dioxide (SO2), a pollutant,

can be removed from the emission of an industrial plant by making it react with

calcium carbonate (CaCO3 and O2). Its chemical reaction is shown below.

2SO2(g) + 2CaCO3(s) + O2(g) → 2CaSO4(s) + 2CO2(g)

1. How many grams of calcium carbonate and oxygen gas are needed to remove 1.25x10^6g of sulfur dioxide from the emission of an industrial plant?

(With solution please)

Answer 1

Molar mass of SO2 = 64.066

1250000/64.066= 19511.13moles

Mole ratio SO2:O2= 2:1

= 19511.13/2= 9755.565moles of O2

Molar mass of O2= 31.998

= 31.998×9755.565=3.122×10⁵g of O₂

Mole ratio SO2: CaCO3 = 2:2

Molar mass of CaCO3= 100.09g/mol

= 100.09×19511.13

= 1.953×10⁶g