Respuesta :
Explanation:
It is given that in 1 liter of solution, there is 300 g of wastewater in [tex]C_{9}N_{2}H_{6}O_{2}[/tex].
Therefore, the reaction equation will be as follows.
[tex]C_{9}N_{2}H_{6}O_{2} + 8O_{2} \rightarrow 2NH_{3} + 9CO_{2}+ 0.H_{2}O[/tex]
For carbonaceous oxygen demand:
Molecular weight of [tex]C_{9}N_{2}H_{6}O_{2}[/tex] = [tex]9(12) + 2(14) + 6(1) + 2(16)[/tex]
= 174 g/mol
or, = [tex]174 \times 10^{3}[/tex] mg/mol (as 1 g = 1000 mg)
So, total moles present in 300 mg of solution will be as follows.
Moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{300 mg}{174 \times 10^{3} mg/mol}[/tex]
= [tex]1.724 \times 10^{-3} mol[/tex]
Hence, for 1 mol of carbonaceous oxygen demand is B mol.
Therefore, for [tex]1.724 \times 10^{-3} mol[/tex] oxygen required is calculated as follows.
[tex]8 \times 1.724 \times 10^{-3} mol[/tex]
= 0.0138 mol
Weight of oxygen required (m) = [tex]0.0138 mol \times 32 g/mol[/tex]
= 0.4414 g
or, = 441.4 mg
This means COD (carbonaceous oxygen demand) is 441.4 mg.
For nitrogenous oxygen demand:
[tex]NH_{3} + 2O_{2} \rightarrow HNO_{3} + H_{2}O[/tex]
For 1 mol of [tex]NH_{3}[/tex] to convert into [tex]HNO_{3}[/tex], oxygen required is 2 mol.
For [tex]2 \times 1.724 \times 10^{-3} mol[/tex] of [tex]NH_{3}[/tex], oxygen is calculated as follows.
[tex]2 \times 2 \times 1.724 \times 10^{-3} mol[/tex]
= 0.006896 mol [tex]O_{2}[/tex]
Mass of [tex]O_{2}[/tex] required is 0.22 g = 220.6 mg
NOD (Nitrogenous oxygen demand) = [tex]\frac{220.6 mg O_{2}}{1 L}[/tex]
Therefore, calculate total biological oxygen demand (BOD) as follows.
= [tex]\frac{(441.4 + 220.6) mg \text{O_{2}}}{1 L}[/tex]
= 662 mg [tex]O_{2}[/tex]/L
