Understanding the Stoichiometry of Water Formation: Determining Oxygen Required

In chemical reactions, stoichiometry is a fundamental concept that helps us determine the quantities of reactants and products involved. The question of how many moles of oxygen are needed to produce 9 grams of water (H2O) is a classic example that involves stoichiometric calculations. Let's delve into the process and calculations step by step.

Understanding the Chemical Reaction

The formation of water can be represented by the following balanced chemical equation:

H2(g) 1/2 O2(g) → H2O(l)

This equation tells us that one mole of hydrogen gas (H2) reacts with one-half mole of oxygen gas (O2) to produce one mole of water (H2O).

Calculating the Molecular Mass of Water

To better understand the stoichiometry, let's first calculate the molecular mass of water. The molecular mass of water, H2O, is:

- Hydrogen (H) has a molecular mass of approximately 1.008 g/mol. - Oxygen (O) has a molecular mass of approximately 16.00 g/mol.

The molecular mass of water (H2O) is therefore:

[ (2 times 1.008) 16.00 18.016 , text{g/mol} approx 18.01 , text{g/mol} ]

Using Stoichiometry to Solve the Problem

Given that the molecular mass of water is approximately 18.01 g/mol, we can now calculate how many moles of water 9 grams represent. This is done using the following formula:

[ text{Moles of water} frac{text{Mass of water}}{text{Molecular mass of water}} ]

Substituting the given values:

[ text{Moles of water} frac{9 , text{g}}{18.01 , text{g/mol}} approx 0.5 , text{mol} ]

According to the balanced equation, to produce 0.5 moles of water, we need one-half mole of O2. Therefore:

[ text{Moles of O2} 0.5 , text{mol} div 2 0.25 , text{mol} ]

Conclusion

In conclusion, to produce 9 grams of water, 0.25 moles of oxygen (O2) are required. This calculation is based on the balanced chemical equation for the formation of water from its constituent elements:

H2(g) 1/2 O2(g) → H2O(l)

It is essential to use a balanced equation and molecular masses to perform such stoichiometric calculations accurately.

Related Keywords

For further reading and exploration, consider researching:

Stoichiometry Oxygen required Molecular mass