2H 2 + O 2 → 2H 2 O From the equation above, calculate the volume of unreacted oxygen gas if a mixture of 50cm…

Analysis

To solve this problem, we use the principle of Gay-Lussac’s Law of Combining Volumes, which states that gases react in simple whole-number ratios by volume under the same conditions of temperature and pressure.

Concept

The balanced chemical equation is: 2H$_2$(g) + O$_2$(g) → 2H$_2$O(g)

From the stoichiometry of the equation:

• 2 volumes of Hydrogen (H$_2$) react with 1 volume of Oxygen (O$_2$).
• This means the ratio of H$_2$ to O$_2$ is 2:1.

Step-by-Step Solution

1. Determine the required volume of Oxygen: Since 2 volumes of H$_2$ require 1 volume of O$_2$, then 50 cm$^3$ of H$_2$ will require: $$\text{Volume of O}_2 \text{ required} = \frac{1}{2} \times 50 \text{ cm}^3 = 25 \text{ cm}^3$$

2. Calculate the unreacted (excess) Oxygen: You were given 75 cm$^3$ of oxygen initially. $$\text{Unreacted O}_2 = \text{Initial Volume} - \text{Volume Used}$$ $$\text{Unreacted O}_2 = 75 \text{ cm}^3 - 25 \text{ cm}^3 = 50 \text{ cm}^3$$

Conclusion

The volume of oxygen remaining after the reaction is 50 cm$^3$.

Exam Tip: Always check the mole ratio from the balanced equation first. If you have two different volumes given for reactants, one is always in "excess." The reactant that runs out first is the "limiting reagent"—always base your calculations on the limiting reagent!