A water sample from a stream with an average flow of 95 000 L/d contains 225 mg/

Question

A water sample from a stream with an average flow of 95 000 L/d contains 225 mg/L of cyanide
waste in the form of sodium cyanide (NaCN). Chlorine can be added to the stream to destroy
this NaCN waste according to the reaction:

2NaCN + 5C12 + 12NaOH ? N2 + Na2 CO3 + 10NaC1 + 6H2 0

Atomic weights are found from a periodic table of the chemical elements to be Na = 22.989
77, C = 12.011, N = 14.0067, Cl = 35.453, O0 = 15.9994, and H = 1.0079.

What is the theoretical minimum amount of chlorine required to destroy the NaCN waste i

Explanation / Answer

given that:

average water flow = 95000 L/d = 95 m3/d

waste in the form of NaCN = 225 mg/L = 225 g/m3

total waste water concentration per day = 95*225 = 21375 g = 21.375 kg.

from the given reaction, we can see that for treating the water, we need 5 moles of chlorine to destroy 2 moles of NaCN.

molecular weight of Cl = 35.453 g/mol

then weight of 5 moles of chlorine = 5*(35.453*2) = 354.53 g

similarily,

molecular weight of Na = 22.98977

C = 12.011

N = 14.0067

then weight of 2 moles of NaCN = 2*(22.98977 +12.011+ 14.0067) = 98.01494 g

that means, we need 354.53 g chlorine to treat 98.01494 g of NaCN waste.

therefore, for treating 1 gram of NaCN required chlorine = 354.53/98.01494

then, for treating 21375 g of NaCN waste, required Chlorine = 21375*(354.53/98.01494) = 77315. 54751 g = 77.31554751 kg.

hence, required chlorine to treat the waste present in the water per day is 77.316 kg.

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