Question

Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4 ( SG = 1.139 ) (SG=1.139) and 60.0 wt% H2SO4 ( SG = 1.498 ) (SG=1.498) are mixed to form a 4.00 molar solution ( SG = 1.213 ) (SG=1.213). Calculate the mass fraction of sulfuric acid in the product solution. Taking 100 kg of the 20% feed solution as a basis, draw and label a flowchart of this process, labeling both masses and volumes, and do the degree-of-freedom analysis. a. Calculate the feed ratio (liters 20% solution/liter 60% solution). b. What feed rate of the 60% solution (L/h) would be required to produce 1250 kg/h of the product?

Answer 1

Answer:

Explanation:

To calculate the mass fraction of sulphuric acid in the product solution, we recall the formula for calculating density which is:

Desnsity = mass/volume

mass = density x volume

Therefore,  Density of product = 1.213 x 1,000 kg/m3

= 1,213 kg/m3 H2SO4 in the product

= (4 mol/L x 98 g/mol x 1,000 L/m3 x 10-3 kg/g)/1,213 kg/m3

mass = 0.323 kg H2SO4/kg

b)

V1 = 100 kg/(1,139 kg/m3 ) = 0.0878 m3 = 87.8 L

Sulfuric acid balance: 100 (0.2) + m2 (0.6) = m3 (0.323) 20 + 0.6 m2 = 0.323 m3 m2 = 0.538 m3 - 33.33

Water balance: 100 (0.8) + m2 (0.4) = m3 (1‐0.323) 80 + 0.4 m2 = 0.677 m3

80 + 0.4 (0.538 m3 - 33.33) = 0.677 m3 80 + 0.2152 m3 - 13.33 = 0.677 m3 66.67 = 0.4618 m3 m3 = 144 kg

V3 = 144 kg/(1,213) kg/m3 = 118.7 m3 m2 = 0.538 (144) - 33.33 = 44.14 kg

V2 = 44 kg/(1,498 kg/m3 ) = 0.0294 m3 = 29.37 L

Feed ratio = 87.8 L (20% solution)/29.37 L (60% solution) = 2.99 L of 20% solution/L of 60% solution

c)

Where,

V2 = 29.4 L required to produce 144 kg of product to produce 1,250 kg of product requires (1,250 x 29.4)/144 = 255 L of 60% solution 100 kg,

V1 (L),  SG = 1.139 0.2 kg H2SO4/kg 0.8 kg H2O/kg m3 kg, V3 (L),  

SG = 1.213 4M H2SO4 (x3) m (1‐x3) kg H2O/kg 2 kg, V2 (L),

SG = 1.498 0.6 kg H2SO4/kg 0.4 kg H2O/kg.