Question

500kgh-1 of a 10% w/w solutionof a compoun X(OH)2 is to be treated using a 5% excess of 20% w/w solution of Na2CO3 to precipitate the X as its carbonate according to the balanced reaction equation: X(OH)2 + Na2CO3 ------> XCO3 + 2NaOH All the solid is filtered off but some solution remains with the solid such that the solution forms 4% w/w of the total mass removed. the filtered liquid product is then neutralised with 2 molar hydrochloric acid according to the reactions NaCO3 + 2HCl -----> 2NaCl + H2O + CO2 NaOH + HCl -----> NaCl + H2O Calculate: 1. the amount (kgh-1) of sodium carbonate solution required. 2. the amount (kgh-1) and composition (%w/w) of the solid product after filtration 3. the amount (kgh-1) and composition (%w/w) of the solid product after filtration. 4. the amount of acid (in litres per hour) required for neutralisation data: Relative atomic mass of X= 40, O=16, H=1, Na= 23, C= 12, Cl= 35.5

Answer 1

weight of solution = 500 *10/100 = 50 kg/hr,    molar mass of X(OH)2= 74

Moles of X(OH)2= 50*1000/74= 675.67 gm moles/hr, According to the reaction, X(OH)2+ Na2CO3---->XCO3+ 2NaOH

moles of Na2CO3 required = 675.67 gmoles/hr. mass of Na2CO3= 675.67*84 = 56756. 28 gm/hr But this is present in 20% solution of Na2CO3. moles of solution = 56756.28/0.2 =283781.4 gm/hr, it is supplied 5% in excess, so mass of Na2CO3 solution = 283781.4*1.05 = 297971gm/hr=297.971 kg/hr

mass of the solution = 3548*31.2 gm/hr =110698 gm/hrr =110.698 kg/hr

mass of Na2CO3 suppiled = 297.971*20/100=59.594 kg/hr, excess = 59.594-56.756 =2.84 kg/hr

moles of solid to be removed= 675.67 moles/hr ( 1mole of X(OH)2 requires 1 mole of Na2CO3 to give one mole of X(OH)2.   mass of the solid = 675.67*100 =67567 gm/hr =67.567 kg/hr, mass of solution with the solid = 67567*4/100 =2703gm/hr =2.703 kg/hr. mass of Na2CLO3 remained = 2.84*1000 gm/hr= 2840 gm/hr= 2840/84 gmoles/hr = 33.81 gmoles/hr, moles of HCl required = 33.81/2= 16.91 gmoles/hr,

molarity of HCl = 2 molar, volume of HCL= 16.91/2= 8.45 L/hr