Understanding the Stoichiometry of Water Formation from Hydrogen and Oxygen

This article delves into the process of forming water (H2O) using hydrogen (H2) and oxygen (O2) in a chemical reaction. We'll explore the concept of stoichiometry through both theoretical and practical approaches. Our goal is to understand whether it is possible to produce 12 grams of water from 8 grams of oxygen and, if not, why.

Stoichiometry Basics

Stoichiometry involves the calculation of reactant proportions in a chemical reaction. To solve this problem, we'll use the balanced chemical equation for the formation of water:

2H2 O2 → 2H2O

This equation tells us that 4 grams of hydrogen (H2) and 32 grams of oxygen (O2) produce 36 grams of water (H2O).

Calculation of Reactant Mole Ratios

Let's calculate the amount of hydrogen required to react with 8 grams of oxygen using the given equation:

Step 1: Write the Chemical Equation

2H2 O2 → 2H2O

Step 2: Determine Molar Masses

Molar mass of H2 2 grams/mole

Molar mass of O2 32 grams/mole

Molar mass of H2O 18 grams/mole

Step 3: Calculate the Number of Moles of Each Compound

For 8 grams of O2 0.25 moles

0.25 * 4 grams of H2 1 gram of H2

0.25 * 32 grams of O2 0.25 * 32 8 grams of O2

Step 4: Determine the Limiting Reactant

To find the limiting reactant, compare the mole ratio of the actual amount of reactants with the mole ratio of the balanced equation:

Needed for 0.25 moles O2: 0.25 * 2 0.5 moles H2

Actually available H2: 1 gram (or 0.5 moles)

Since the actual amount of H2 is exactly what is needed, neither reactant is in excess.

Step 5: Calculate the Amount of Water Produced

Since 0.25 moles of O2 will react with 0.5 moles of H2, we can produce 0.25 moles of H2O:

0.25 moles H2O * 18 grams/mole 4.5 grams of H2O

Conclusion and Additional Insights

The provided solution is accurate, as only 4.5 grams of water can be formed from 8 grams of oxygen, not 12 grams. This is because the amount of hydrogen provided (1 gram, or 0.5 moles) is just enough to react with the oxygen, leaving no excess hydrogen remaining.

For a faster method, one can use a simplified formula to estimate the mass of water produced:

Mass of water Mass of oxygen * (Molar mass of H2O / Molar mass of O2)

This equation gives us:

2 grams of O2 * (18.015 g/mol / 15.999 g/mol) 2.252 grams of H2O

Thus, 2.252 grams of water will be formed.

This approach not only saves time but also confirms that the mass is conserved in the reaction, as all the oxygen is converted into water.

Additional Resources

For more detailed understanding and additional practice problems, you can refer to the following resources:

Comprehensive Guide on Hydrogen and Oxygen Reactions Stoichiometry Summary from BC Open Text Interactive Stoichiometry Exercises