The reaction you've described is the oxidation of ethanol (CH3CH2OH) to form water (H2O) and acetic acid (CH3COOH) in the presence of oxygen (O2) from the air. The balanced chemical equation for this reaction is:
C2H5OH + O2 -> H2O + CH3COOH
From the balanced equation, it's clear that 1 mole of oxygen gas (O2) reacts to produce 1 mole of water (H2O). The molar mass of oxygen gas (O2) is approximately 32 g/mol, and the molar mass of water (H2O) is approximately 18 g/mol.
Given that the reaction involves 3.7 g of oxygen gas, you can calculate the amount of water produced using the molar ratios:
Amount of oxygen gas (in moles) = 3.7 g / 32 g/mol ≈ 0.1156 mol
Since the reaction produces an equal amount of water (1:1 molar ratio), the amount of water produced is also approximately 0.1156 moles.
Now, calculate the mass of water produced:
Mass of water = Amount of water (in moles) × Molar mass of water
Mass of water = 0.1156 mol × 18 g/mol ≈ 2.08 g
Rounding to two significant digits, the mass of water produced is approximately 2.1 g.