Question

A plant makes liquid carbon dioxide by treating a limestone with commercial sulfuric acid. The limestone analyzes 68.1% calcium carbonate, 27.1% magnesium carbonate and the rest silica. The acid used was 88.3% sulfuric acid with rest being water. Calculate the pounds of acid required per ton of limestone treated.

Answer 1

Composition of limestone = CaCO3 + MgCO3 + SiO2

Percentage of CaCO3 = 68.1%

CaCO3 + H2SO4 -----> CaSO4 + H2O + CO2

1 mole of CaCO3 requires 1 mole of H2SO4

1 mole of lime stone contains 68.1% of CaCO3

so, 0.681 moles of CaCO3 requires 0.681 moles of H2SO4

But it is given that only 88.3% H2SO4 is present in 1 unit of commercial H2SO4

Hence H2SO4 required = 0.681*100/88.3 = 0.7712 moles of commercial H2SO4

Percentage of MgCO3 = 27.1%

MgCO3 + H2SO4 --------> MgSO4 + H2O + CO2

1mole of MgCO3 requires 1 mole of H2SO4

1 mole of lime stone contains 27.1% of MgCO3

so, 0.271 moles of CaCO3 requires 0.271 moles of H2SO4

But it is given that only 88.3% H2SO4 is present in 1 unit of commercial H2SO4

Hence H2SO4 required = 0.271*100/88.3 = 0.307 moles of commercial H2SO4

Lime stone composition = 68.1% CaCO3 + 27.1% MgCo3 + 4.8% of SiO2

so 1 mole of Lime stone requires (0.7712+0.307) moles of Commercial H2SO4

1 mole of Lime stone requires 1.0782 moles of Commercial H2SO4

0.681 moles of CaCO3 + 0.271 moles of MgCO3 + 0.048 moles of SiO2 requires 1.0782 moles of Commercial H2SO4

(0.681*100)+(0.271*84.3)+(0.048*60) g of lime stone requires 1.0783*98 g of Commercial H2SO4

93.8253 g of lime stone requires 105.6636 g of commercial H2SO4

we have 1 ton = 907185 g

907185 g of lime stone requires 105.6636*907185/93.8253 g of commercial H2SO4

                               = 1021648.031 g of commercial H2SO4 is needed.

mass of H2SO4 in pounds = 1021648.031g * 1 pound/453.592 g

Required mass of Commercial H2SO4 = 2252.35 pounds