1. Two aqueous sulfuric acid solutions containing 15.0 wt% H2SO, sG-1.139, and 7

Question

1. Two aqueous sulfuric acid solutions containing 15.0 wt% H2SO, sG-1.139, and 70.0 wt% H2SO4, SG-1.498, are mixed to form a 5.00 molar solution, SG-1.213. a) Calculate the mass fraction of sulfuric acid in the product solution b) Taking 100 kg/hr of the 15% feed solution as a basis, draw and label a flowchart of this process, then calculate the mass flow rates c) What feed rate of the 70% solution would be required to produce 1250 kg/hr, in Lhr? Cannot take a basis of 100 kg/hr for the 15% feed solution.

Explanation / Answer

Part a

Final product

Specific gravity of final solution = 1.213

Density of final solution = Specific gravity x density of water

= 1.213 x 1000 kg/m3

= 1213 kg/m3

Molarity of final solution = 5 mol/L

Mass concentration of H2SO4 in final solution

= 5 mol/L x 98 g/mol

= 490 g/L x (1kg/m3)/(1g/L)

= 490 kg/m3

Mass fraction of H2SO4 in final solution

= 490/1213

= 0.4040

Part b

Volume of 15 wt% solution = mass/density

= (100 kg/hr) / (1139 kg/m3)

= 0.0878 m3

H2SO4 balance

100 x 0.15 + m2 x 0.70 = m3 x 0.404

Multiply by 0.30

4.5 + 0.21m2 = 0.1212 m3

Water balance

100 x 0.85 + m2 x 0.30 = m3 x 0.596

Multiply by 0.70

59.5 + 0.21 m2 = 0.4172 m3

Solve these equations simultaneously by subtracting

55 = 0.296 m3

m3 = 185.81 kg

m2 = 85.81 kg

Part c

V2 = m2/d2 = (85.81 kg)/(1498 kg/m3) = 0.05728 m3

= 57.28 L

V3 = m3/d3 = (185.81 kg)/(1139 kg/m3) = 0.1631 m3

= 163.1 L

To produce 70% solution

Feed rate = (1250 kg/h x 57.28 L)/85.81kg

= 834.4 L/h

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