Calculating the Mass of Oxygen Required to React Completely with Hydrogen to Form Water

Gaseous hydrogen reacts with gaseous oxygen to produce water, following the well-known chemical reaction:

2H2(g) O2(g) → 2H2O(l)

This balanced equation indicates that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water. As a result, this stoichiometric relationship can be used to determine the mass of oxygen required when an exact amount of hydrogen is provided.

Molecular Stoichiometry and Mass Calculation

Given that 1 gram of hydrogen reacts with 8 grams of oxygen to form 9 grams of water, we can extrapolate this ratio to find the mass of oxygen required for a larger quantity of hydrogen.

Example 1: If 3 grams of hydrogen are provided, then the mass of oxygen required to react completely can be calculated as follows:

Mass of oxygen 3g × 8 24 grams

For a more precise calculation using the stoichiometric ratios:

2 moles of H2 require 1 mole of O2 Expressing in liters for a gaseous reaction: 2 L of H2 require 1 L of O2

Given 3.6 L of H2, calculate the volume of O2 required:

X 3.6 L H2 / 2 L H2 × 1 L O2 1.8 L O2

These calculations are essential in determining the necessary amounts for both reactants.

Interpreting the Reaction Equation

The chemical equation for the formation of water from hydrogen and oxygen can be expressed as:

H2(g) 0.5O2(g) → H2O(g)

This equation can be interpreted as 2 grams of dihydrogen (H2) reacting with 16 grams of dioxygen (O2) to yield 18 grams of water (H2O).

To find the mass of dioxygen required to produce 3.5 moles of water, we use the molar masses and stoichiometric relationships:

0.5 × 3.50 mol × 32.0 g/mol 56.0 g O2

This calculation shows that 56 grams of dioxygen are required to produce 3.5 moles of water.

Stoichiometric Calculation and Balanced Equation

A balanced chemical equation is key for accurate stoichiometric calculations:

2H2(g) O2(g) → 2H2O(l)

From the balanced equation, we know that 2 moles of hydrogen gas (H2) will react with 1 mole of oxygen gas (O2) to produce 2 moles of water (H2O).

Given 3.5 moles of hydrogen gas:

3.5 moles H2 / 2 moles H2 × 1 mole O2 1.75 moles O2

Since the molar mass of O2 is 32 g/mol, the mass of 1.75 moles of O2 can be calculated as:

1.75 moles × 32 g/mol 56 grams

This shows that 56 grams of oxygen are required to react completely with 3.5 moles of hydrogen to produce water.

Conclusion

The mass of oxygen required to react completely with a given amount of hydrogen can be calculated using the principles of stoichiometry and the balanced chemical equation. By understanding these concepts, one can ensure the proper proportions of reactants are maintained for efficient and safe chemical reactions.