Introduction:
In order to determine the amount of hydrogen that would diffuse through a very narrow hole under identical conditions, we need to consider the molar ratios of oxygen and hydrogen in a chemical reaction. By doing so, we can apply the concept of stoichiometry to calculate the amount of hydrogen.

Stoichiometry and Molar Ratios:
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the principle of the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction, only rearranged.

The molar ratio is the ratio of the coefficients of reactants and products in a balanced chemical equation. It allows us to determine the amount of one substance based on the amount of another substance involved in the reaction.

The Balanced Equation for the Reaction:
To determine the molar ratio between oxygen and hydrogen, we need to consider the balanced equation for the reaction. In the case of oxygen and hydrogen, they react to form water according to the following equation:

2H₂ + O₂ → 2H₂O

From this equation, we can deduce that for every 1 mole of oxygen, 2 moles of hydrogen are required to react completely.

Calculating the Amount of Hydrogen:
Given that 4 g of oxygen has diffused through the narrow hole, we need to determine the number of moles of oxygen. To do this, we use the molar mass of oxygen, which is approximately 32 g/mol.

Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
= 4 g / 32 g/mol
= 0.125 mol

Since the molar ratio between oxygen and hydrogen is 1:2, the number of moles of hydrogen can be calculated as follows:

Number of moles of hydrogen = Number of moles of oxygen × 2
= 0.125 mol × 2
= 0.25 mol

Therefore, under identical conditions, 0.25 moles of hydrogen would have diffused through the narrow hole.

Conclusion:
Applying the principles of stoichiometry and the molar ratio between oxygen and hydrogen, we have determined that if 4 g of oxygen diffuse through a very narrow hole, 0.25 moles of hydrogen would also diffuse under identical conditions.

16 gm