Question

Show that 1 L of water saturated with oxygen at 25°C is capable of completely oxidizing 8.2 mg of polymeric CH2O

Answer 1

Solubility of O2 in water at 25 oC = 8.7 mg/L

So, 1 L of water saturated with oxygen contains 8.7 mg of oxygen.

Mass of CH2O present = 8.2 mg = 0.0082 g

Molar mass of CH2O = 30 g/mol

Moles of CH2O = mass/molar mass = 0.0082/30 = 0.00027 mol

Mass of O2 present = 8.7 mg = 0.0087 g

Molar mass of O2 = 32 g/mol

Moles of O2 = mass/molar mass = 0.0087/32 = 0.00027 mol

For oxidation of CH2O:

CH₂O + O₂ ----> CO₂ + H₂O

So, 1 mol of CH2O requires 1 mol of O2 for complete oxidation.

Thus 0.00027 mol of CH2O will need 0.00027 mol of O2 for complete oxidation.

Thus, it will be capable of oxidising polymeric CH2O.