Concept:This is an application of Eudiometry. Under standard conditions, gases react in simple whole-number volume ratios. The decrease in volume is attributed to the gaseous reactants completely converting into liquid water, whose gaseous volume essentially disappears.
Formula:$$ 2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_2\text{O}_{(l)} $$
Solution:- The balanced equation states that 2 volumes of Hydrogen react with 1 volume of Oxygen.
- We are given \( 50\text{cm}^3 \) of \( \text{H}_2 \) and \( 10\text{cm}^3 \) of \( \text{O}_2 \).
- According to the 2:1 ratio, the \( 10\text{cm}^3 \) of Oxygen will strictly react with only \( 20\text{cm}^3 \) of Hydrogen. (Oxygen is the limiting reactant).
- Total volume of reactants consumed = \( 10\text{cm}^3 \text{ (O}_2) + 20\text{cm}^3 \text{ (H}_2) = 30\text{cm}^3 \).
- Since the product (water) is a liquid under normal room conditions, its volume is negligible compared to the original gases.
- Therefore, the total decrease in gaseous volume is precisely the volume of the gases consumed: \( 30\text{cm}^3 \).
Why other options are incorrect:- \( 10\text{cm}^3 \) or \( 20\text{cm}^3 \): These are merely the independent volumes of the individual reactants consumed, not their combined total.
- \( 15\text{cm}^3 \): Has no stoichiometric basis in a 2:1 ratio reaction.
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