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A tank is filled to 80% of its capacity with a saturated solution of MgSO4. The solution contains 1800 kg of MgSO, and is kept at 360 K. The tank is opened and the solution cools to 285 K forming MgSO4.8H2O crystals. Calculate the resultant yield of crystals and the capacity (volume) of the tank. Use the information below: • Density of the solution is 1900 kg/m • Solubility of M£SO4 per 100 parts of water by mass is 50 at 360 K and 33 at 285 K. • Assume there is negligible evaporation • Molecular weight of MgSO, is 120 kg/kmol • Molecular weight of H3O is 18 kg/kmol |C1 – C2(1 – E)| li- c2 (R – 1) Y = RW,

A tank is filled to 80% of its capacity with a saturated solution of MgSO4. The solution contains 1800 kg of MgSO, and is kept at 360 K. The tank is opened and the solution cools to 285 K forming MgSO4.8H2O crystals. Calculate the resultant yield of crystals and the capacity (volume) of the tank. Use the information below: • Density of the solution is 1900 kg/m • Solubility of M£SO4 per 100 parts of water by mass is 50 at 360 K and 33 at 285 K. • Assume there is negligible evaporation • Molecular weight of MgSO, is 120 kg/kmol • Molecular weight of H3O is 18 kg/kmol |C1 – C2(1 – E)| li- c2 (R – 1) Y = RW,

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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A tank is filled to 80% of its capacity with a saturated solution of MgSO4. The solution contains 1800 kg of MgSO, and is kept at 360 K. The tank is opened and the solution cools to 285 K forming MgSO4.8H2O crystals. Calculate the resultant yield of crystals and the capacity (volume) of the tank. Use the information below: • Density of the solution is 1900 kg/m • Solubility of M£SO4 per 100 parts of water by mass is 50 at 360 K and 33 at 285 K. • Assume there is negligible evaporation • Molecular weight of MgSO, is 120 kg/kmol • Molecular weight of H3O is 18 kg/kmol |C1 – C2(1 – E)| li- c2 (R – 1) Y = RW,
Transcribed Image Text:A tank is filled to 80% of its capacity with a saturated solution of MgSO4. The solution contains 1800 kg of MgSO, and is kept at 360 K. The tank is opened and the solution cools to 285 K forming MgSO4.8H2O crystals. Calculate the resultant yield of crystals and the capacity (volume) of the tank. Use the information below: • Density of the solution is 1900 kg/m • Solubility of M£SO4 per 100 parts of water by mass is 50 at 360 K and 33 at 285 K. • Assume there is negligible evaporation • Molecular weight of MgSO, is 120 kg/kmol • Molecular weight of H3O is 18 kg/kmol |C1 – C2(1 – E)| li- c2 (R – 1) Y = RW,
Expert Solution
Step 1

Given data:

Amount of MgSO4 in the feed = 1800 kg

Temperature of feed = 360 k

Temperature of mother liquor = 285 K

Density of the solution = 1900 kg/m3

Solubility of MgSO4 at 360 K = 50/100 parts water

Solubility of MgSO4 at 285 K = 33 /100 parts water

Molar mass of MgSO4 = 120 kg/kmol

Molar mass of H2O = 18 kg/kmol

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