The reaction of lithium metal and water to form lithium hydroxide and hydrogen gas is represented by the following balanced chemical equation: 2Li (s) + 2H2O (l) → 2LiOH (aq) + H2 (g) When Li is mixed with excess water, 0.30 mol of H2 gas is isolated in the laboratory. If this reaction occurs to give an 85% yield of H2, how many moles of Li reacted?
The correct Answer and Explanation is:
Step 1: Understanding the Reaction
The balanced chemical equation is:
2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)2Li (s) + 2H_2O (l) \rightarrow 2LiOH (aq) + H_2 (g)
From this equation, 2 moles of lithium (Li) produce 1 mole of hydrogen gas (H₂).
Step 2: Relating Yield to Theoretical Values
You are given that 0.30 mol of H₂ was isolated at an 85% yield. Theoretical yield means how much hydrogen would be produced if the reaction were 100% efficient.
Using the formula:
Theoretical yield=Actual yieldPercent yield\text{Theoretical yield} = \frac{\text{Actual yield}}{\text{Percent yield}}
Theoretical yield of H₂=0.300.85=0.353 mol\text{Theoretical yield of H₂} = \frac{0.30}{0.85} = 0.353 \text{ mol}
Step 3: Calculating the Required Moles of Lithium
Since 2 moles of lithium produce 1 mole of H₂, the number of moles of lithium required for the theoretical yield is:
Moles of Li=2×Moles of H₂\text{Moles of Li} = 2 \times \text{Moles of H₂}
Moles of Li=2×0.353=0.706 mol\text{Moles of Li} = 2 \times 0.353 = 0.706 \text{ mol}
Conclusion
So, 0.71 moles of Li (rounded to two decimal places) reacted in this experiment.
Explanation
Lithium, being highly reactive, reacts with water to form lithium hydroxide and hydrogen gas. The key to solving this problem was understanding stoichiometry, which allows us to relate reactants and products in a chemical reaction. Since the reaction does not proceed at full efficiency (85% yield), we first calculated the theoretical hydrogen yield before using stoichiometric relationships to determine the lithium needed.
Chemical reactions often do not reach 100% yield due to loss of material, side reactions, or incomplete conversions. This is why adjusting for percent yield is essential. Knowing how to determine reactant quantities accurately helps in optimizing real-world chemical processes, such as those used in battery production and laboratory synthesis.
