Problem Set 5#14

Mark Lorenz CapiliArlyn Castillo Jerina Joy PangilinanJustine Noel Sison

An alkaline solution of sodium carbonate made up of 0.594% NaOH,14.88%

Na2CO3 and 84.53% Water is causticized by the addition of partly slaked

lime. The lime contains CaCO3, CaO and Ca(OH) 2. The mass obtained

from the causticization has the following analysis: 13.48% CaCO3, 0.28%

Ca(OH) 2 , 0.61% Na2CO3, 10.36% NaOH, and the rest is water. The

reactions are:

CaO + H2O = Ca(OH) 2

Na2CO3 + Ca(OH) 2 = 2NaOH + CaCO3

On a basis of 100 kg of the causticized mass, calculate:

a) Wt. and composition of the lime

b) Wt. of the alkaline liquor

c) % excess reactant

d) Degree of completion of the overall reaction

Lime (y)

•CaO• CaCO3

•Ca(OH) 2

Alkaline liquor (x)

•0.594% NaOH•14.88% Na2CO3

•84.53% H2O

Causticized mass (100 kg)

•13.48% CaCO3

•0.28% Ca(OH)2

•0.61% Na2CO3

•10.36% NaOH

•75.27% H2O Basis: 100 kg causticized mass• 13.48 kg CaCO3

• 0.28 kg Ca(OH)2

• 0.61 kg Na2CO3

• 10.36 kg NaOH

• 75.27 kg H2O

Na balance:

(X) (0.00594) kg NaOH x 23 kg Na + (X) (0.1488) kg Na2CO3 x 2(23) kg Na

40kg NaOH 106 kg Na2CO3

= 0.61 kg Na2CO3 x 2(23)kg Na + 10.36 kg NaOH x 23 kg Na

106 kg Na2CO3 40 kg NaOH

OMB: x + y = 100X = 91.5 kg Alkaline liquor

Y= 100 kg – 91.5 kg= 8.5 kg Lime

CaO + H2O = Ca(OH) 2

Supplied (kg) 6.4 77.34 Unreacted (kg) 0 75.27 Reacted (kg) 6.4 2.0 Produced (kg) 8.52

Amt. H2O reacted = (91.5) (0.8453) kg H2O (supplied) – 75.27 kg H2O (unreacted)

= 77.34 kg H2O (supplied) – 75.27 kg H2O (unreacted)

= 2.07 kg H2O (reacted)

Amt. CaO reacted = 2.07 kg H2O x 56.08 kg CaO

18 kg H2O

= 6.4 kg CaO (reacted)

Amt. Ca(OH) 2 produced = 2.07 kg H2O x 74.09 kg Ca(OH) 2

18 kg H2O

= 8.52 kg Ca(OH) 2 (produced)

Na2CO3 + Ca(OH) 2 = 2NaOH + CaCO3

Supplied (kg) 13.62 8.52 + a 0.54351 b Unreacted (kg) 0.61 0.28Reacted (kg) 13.01 9.09Produced (kg) 9.82 12.29Total product (kg) 10.36 13.48

Amt. Na2CO3 reacted = (91.5 kg alkaline liquor) (0.1488) – 0.61 kg Na2CO3 (unreacted)

= 13.62 kg Na2CO3 (supplied) – 0.61 kg Na2CO3 (unreacted)= 13.01 kg Na2CO3 (reacted)

Amt. Ca(OH) 2 reacted = 13.01 kg Na2CO3 x 74.09 kg Ca(OH) 2

106 kg Na2CO3

= 9.09 kg Ca(OH) 2 (reacted)

Amt. NaOH produced = 13.01 kg Na2CO3 x 2(40)kg NaOH

106 kg Na2CO3

= 9.81 kg NaOH (produced)

Amt. CaCO3 produced = 13.01 kg Na2CO3 x 100.08 kg CaCO3

106 kg Na2CO3

= 12.29 kg CaCO3 (produced)

Amt. Ca(OH) 2 supplied = amt. Ca(OH) 2 reacted + amt. Ca(OH) 2 unreacted

8.52 kg + a = 9.09 kg + 0.28 kg

a = 0.85 kg Ca(OH) 2 entering

Total CaCO3 produced = CaCO3 produced in the 2nd rxn + CaCO3 entering

13.48 kg = 12.29 kg + b

b = 1.19 kg CaCO3 entering

The composition of the following in the lime are:

Wt. of lime =8.5 kg

CaO = (6.46kg/ 8.5 kg) x 100% = 76% CaO

CaCO3 = (0.85kg/ 8.5kg) x 100% = 10% CaCO3

Ca(OH) 2 = (1.19kg/ 8.5kg) x 100% = 14% Ca(OH) 2

c) % excess reactant

From the reaction shown below, to determine the limiting reactant:

Na2CO3 + Ca(OH) 2 = 2NaOH + CaCO3

Assume Ca(OH) 2 as limiting reactant:

9.37 kg Ca(OH) 2 x 106 kg Na2CO3 = 13.4 kg Na2CO3 < 13.62 kg

74.09 kg Ca(OH) 2

therefore, Ca(OH) 2 is the limiting reactant

% excess Na2CO3 = 13.62 kg Na2CO3 (supplied) - 13.4 kg Na2CO3 (theoretical) x 100%

13.4 kg Na2CO3 (theoretical)

= 1.64 % excess Na2CO3

D) Degree of Completion

Degree of Completion = 9.09 kg Ca(OH) 2 (reacted) x 100%9.37 kg Ca(OH) 2 (supplied)

= 97.01 %

Thank you for Listening!