please help Crystalline sodium Bicarbonate may be decomposed to Sodium Carbonate

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Crystalline sodium Bicarbonate may be decomposed to Sodium Carbonate (Light Ash) according to the overall reaction In reality the reaction proceeds via a number of intermediate co-ordination complexes of Sodium Bicarbonate, Sodium Carbonate and water. An internally heated reacting drser (Secheur) produces 27000 kg.h^-1 of dried solid that contains 97% w/w sodium carbonate and 3% w/w of the co-ordination complex 3NaHCO_3 - Na_2CO_3. H_2O which may be considered as a mixture of un-reacted Sodium Bicarbonate, Sodium Carbonate and un-evaporated Water. The drier is fed with Sodium Bicarbonate crystals that contain 13% water. Calculate The feed rate of wet solid to the drier The flow rate and composition of the effulent gas stream. Data Relative molar masses

Explanation / Answer

a) Feed rate of wet solid to the drier: Mass of sodium carbonate: mNa2CO3 = (97/100) x 27000 kg/h mNa2CO3 = 26190 kg/h In mol, nNa2CO3 = 26190 kg/h/0.106 kg/mol nNa2CO3 = 247075.472 mol/h Mass of residue: mresidue = 27000 kg/h – 26190 kg/h mresidue = 810 kg/h In mol, first calculate average molar mass: Mresidue = (3/5).84 g/mol + (1/5).106 g/mol + (1/5).18 g/mol Mresidue = 75.2 g/mol nresidue = 810 kg/h/0.0752 kg/mol nresidue = 10771.28 mol/h (a fifth of this is water) nresidue = (4/5).10771.28 mol/h nresidue = 8617.024 mol/h Total moles of solid that exits the drier: ntotal = nNa2CO3 + nresidue ntotal = 255692.5 mol/h According to the chemical equation, the feed of dry NaHCO3 is the double of ntotal: Moles of NaHCO3 in the feed: nNaHCO3 = 2 x 255692.5 mol/h nNaHCO3 = 511385 mol/h Mass of NaHCO3 in the feed: mNaHCO3 = 511385 mol/h . 0.084 kg/mol mNaHCO3 = 42956.34 kg/h Feed rate of wet solid to the drier: F = mNaHCO3/(100-13/100) F = 49375.1 kg/h* *this result differs in 1.2% to the answer given due to approximation in previous numbers. b) Flow rate and composition of effluent gas stream: Moles of NaHCO3 in the feed: nNaHCO3 = 511385 mol/h According to the chemical equation: Moles of CO2 in the stream: nCO2 = (1/2).511385 mol/h nCO2 = 255692.5 mol/h Mass of CO2: mCO2 = 255692.5 mol/h . 0.044 kg/mol mCO2 = 11250.47 kg/h Moles of H2O in the stream from reaction: nH2O = nCO2 nH2O = 255692.5 mol/h Mass of H2O from reaction: mH2O = 255692.5 mol/h . 0.018 kg/mol mH2O = 4602.47 kg/h (we have to consider the 13% water from the wet feed) mH2O = 4602.47 kg/h + (49375.1 kg/h).(13/100) mH2O = 11021.23 kg/h Total mass: T = mCO2 + mH2O T = 22271.7 kg/h Composition of the stream %CO2 = (mCO2/T).100 %CO2 = 50.51 %H2O = 100 – 50.51 %H2O = 49.49

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