500kgh-1 of a 10% w/w solutionof a compoun X(OH)2 is to be treated using a 5% ex

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



Any working out would be a great help. Thanks

Explanation / Answer

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