Question

A gold-plating business produces a waste with cyanide (CN–), which is a poisonous material. The cyanide can be oxidized in a two step process as shown below.

Reaction 1:

Cyanide to Cyanate: NaCN + NaOCl → NaCNO + NaCl

Reaction 2: Cyanate Destruction to Gases 2 NaCNO + 3 NaOCl + H2O → 3 NaCl + N2 + 2 NaHCO3

How much bleach (NaOCl) in kg/yr is needed to safety destroy cyanide at the gold plating business if the waste stream is 5,000 L/d and the cyanide concentration is 2,000 mg/L of CN– ions?

Answer 1

Amount of waste produced = 5000 L/d

Amount of waste produced per year = 5000 X 365

=1825000 L

Amount of cyanide in the waste = 2000mg/L =2g/L

Therefore,

Amount of cyanide produced per year = 1825000 X 2

= 3650000g

In the first step,

NaCN + NaOCl → NaCNO + NaCl

1 mole of NaOCl is required

In the next step,ie,

2 NaCNO + 3 NaOCl + H2O → 3 NaCl + N2 + 2 NaHCO3

For 1 mole of NaCNO, 3/2 = 1.5 mole of NaOCl is required.

Therefore, from the two processes,

Number of moles of NaOCl required = 1.5 + 1 = 2.5

Molecular weight of NaOCl = 74.44g

Amount of NaOCl required = 74.44 X 2.5 = 186.1g

molecular weight of NaCN = 49

i,e. 49g of NaCN requires 186.1g of NaOCl

For 1g NaCN,

186.1/49 = 3.79g NaOCl is required.

Therefore NaOCl required per year = 3.79 X 3650000

= 13833500g

=13833.5 Kg

Thus 13833.5 kg of NaOClis required per year.