2H2(g) + O2(g) → 2H2 O(g) In the reaction above, what volume of hydrogen would be left over when 300 cm3 of…

Analysis

To solve this problem, we use 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: 2H₂(g) + O₂(g) → 2H₂O(g)

From the stoichiometry of the equation:

• 2 volumes of Hydrogen (H₂) react with 1 volume of Oxygen (O₂).
• This means for every 1 cm³ of O₂, you need 2 cm³ of H₂.

Step-by-Step Solution

1. Determine the required volume of Hydrogen: We are given 300 cm³ of Oxygen. According to the ratio (2:1), the volume of Hydrogen needed to react completely with 300 cm³ of Oxygen is: Volume of H₂ needed = 2 × Volume of O₂ Volume of H₂ needed = 2 × 300 cm³ = 600 cm³

2. Calculate the volume of Hydrogen left over: We started with 1000 cm³ of Hydrogen. Volume of H₂ left = (Initial volume of H₂) - (Volume of H₂ used) Volume of H₂ left = 1000 cm³ - 600 cm³ = 400 cm³

Conclusion

Since 600 cm³ of Hydrogen is consumed to react with the 300 cm³ of Oxygen, 400 cm³ of Hydrogen remains unreacted.

Exam Tip: Always check the stoichiometry (the coefficients in the balanced equation) first. If you are given two volumes, one of them is usually the "limiting reactant." In this case, Oxygen was the limiting reactant because it ran out first, leaving the excess Hydrogen behind.