Stoichiometric Analysis of Water Formation from Hydrogen and Oxygen

Stoichiometric Analysis of Water Formation from Hydrogen and Oxygen

Determining how much water can be produced from a given amount of hydrogen and oxygen involves a step-by-step approach using stoichiometry based on the balanced chemical equation for water formation. This article will walk you through the process, highlighting the importance of both the balanced equation and the concept of the limiting reactant.

Chemical Equation and Molar Masses

The balanced chemical equation for the formation of water from hydrogen and oxygen is:

2 H2 O2 → 2 H2O

Here, the molar masses of the reactants are:

Hydrogen (H): 2 g/mol Oxygen (O): 16 g/mol Water (H2O): 18 g/mol

Calculating Moles of Reactants

Let's start by calculating the moles of hydrogen and oxygen available in the given amounts:

Moles of hydrogen (H2):

2.00 g ÷ 2.02 g/mol ≈ 0.9901 mol

Moles of oxygen (O2):

15.87 g ÷ 32.00 g/mol ≈ 0.4966 mol

Determining the Limiting Reactant

To determine the limiting reactant, we need to compare the mole ratio of the reactants with the stoichiometric ratio given by the balanced equation. From the balanced equation:

2 moles of H2 react with 1 mole of O2

Our calculated moles give us a ratio of 0.9901 moles of H2 to approximately 0.4966 moles of O2. We need to see if this ratio matches the stoichiometric ratio.

Required moles of O2 for 0.9901 moles of H2:

0.9901 moles H2 ÷ 2 0.49505 moles O2

Since we have 0.4966 moles of O2, which is more than the 0.49505 moles required, hydrogen (H2) is the limiting reactant.

Calculating the Amount of Water Produced

Given that hydrogen is the limiting reactant, the amount of water produced will be based on the moles of H2 consumed. According to the balanced equation:

2 moles of H2 produce 2 moles of H2O

Therefore, the moles of H2O produced will be the same as the moles of H2 consumed:

Moles of H2O 0.9901 mol H2

Using the molar mass of water (18 g/mol), we can convert the moles of H2O to grams:

Mass of H2O 0.9901 mol × 18 g/mol ≈ 17.83 g

Conclusion

Therefore, when 2.00 g of hydrogen (H2) is completely reacted with 15.87 g of oxygen (O2) to produce water (H2O), the amount of water produced will be approximately 17.83 grams.