Question

Determine the carbonaceous and nitrogenous oxygen demand in mg/L for a 1 L solution containing 300 mg of a wastewater represented by the formula CN2H602 (N is converted to NH3 in the first step) Afterwards calculate the COD of the solution

Answer 1

Step-by-step explanation:

It is given that in 1 liter of solution, there is 300 g of wastewater in
`C_(9)N_(2)H_(6)O_(2)`.

Therefore, the reaction equation will be as follows.

`C_(9)N_(2)H_(6)O_(2) + 8O_(2) \rightarrow 2NH_(3) + 9CO_(2)+ 0.H_(2)O`

For carbonaceous oxygen demand:

Molecular weight of
`C_(9)N_(2)H_(6)O_(2)` =
`9(12) + 2(14) + 6(1) + 2(16)`

= 174 g/mol

or, =
`174 * 10^(3)` mg/mol (as 1 g = 1000 mg)

So, total moles present in 300 mg of solution will be as follows.

Moles =
`\frac{mass}{\text{molar mass}}`

=
`(300 mg)/(174 * 10^(3) mg/mol)`

=
`1.724 * 10^(-3) mol`

Hence, for 1 mol of carbonaceous oxygen demand is B mol.

Therefore, for
`1.724 * 10^(-3) mol` oxygen required is calculated as follows.

`8 * 1.724 * 10^(-3) mol`

= 0.0138 mol

Weight of oxygen required (m) =
`0.0138 mol * 32 g/mol`

= 0.4414 g

or, = 441.4 mg

This means COD (carbonaceous oxygen demand) is 441.4 mg.

For nitrogenous oxygen demand:

`NH_(3) + 2O_(2) \rightarrow HNO_(3) + H_(2)O`

For 1 mol of
`NH_(3)` to convert into
`HNO_(3)`, oxygen required is 2 mol.

For
`2 * 1.724 * 10^(-3) mol` of
`NH_(3)`, oxygen is calculated as follows.

`2 * 2 * 1.724 * 10^(-3) mol`

= 0.006896 mol
`O_(2)`

Mass of
`O_(2)` required is 0.22 g = 220.6 mg

NOD (Nitrogenous oxygen demand) =
`(220.6 mg O_(2))/(1 L)`

Therefore, calculate total biological oxygen demand (BOD) as follows.

=
`\frac{(441.4 + 220.6) mg \text{O_(2)}}{1 L}`

= 662 mg
`O_(2)`/L