The reaction of lithium metal and water to form lithium hydroxide and hydrogen gas is represented by the following balanced equation: 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) When Li is mixed with excess water, 0.30 mol 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 reaction of lithium metal and water to form lithium hydroxide and hydrogen gas is represented by the following balanced equation: 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) When Li is mixed with excess water, 0.30 mol 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:
To solve this problem, we use stoichiometry and the concept of percent yield.
Step 1: Write the balanced chemical equation
$$
2Li(s) + 2H_2O(l) \rightarrow 2LiOH(aq) + H_2(g)
$$
This equation tells us that 2 moles of lithium (Li) react to form 1 mole of hydrogen gas (H₂).
Step 2: Understand percent yield
You are told that 0.30 mol of H₂ was isolated with an 85% yield. This means that 0.30 mol is only 85% of the amount of H₂ that would have been produced if the reaction had gone to 100% completion (theoretical yield).
Use the percent yield formula:
$$
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100
$$
Rearrange to solve for theoretical yield:
$$
\text{Theoretical yield} = \frac{\text{Actual yield}}{\text{Percent yield}} = \frac{0.30 \, \text{mol}}{0.85} \approx 0.3529 \, \text{mol of } H_2
$$
Step 3: Use stoichiometry to find moles of Li
From the balanced equation:
$$
2 \, \text{mol Li} : 1 \, \text{mol } H_2
$$
So to make 0.3529 mol of H₂, we need:
$$
2 \times 0.3529 = 0.7058 \, \text{mol of Li}
$$
✅ Final Answer:
$$
\boxed{0.71 \, \text{mol of Li (rounded to two decimal places)}}
$$
Explanation (300+ words):
This problem involves applying basic stoichiometry and yield concepts to a chemical reaction. Lithium reacts with water to produce lithium hydroxide and hydrogen gas. The balanced equation, $2Li + 2H_2O \rightarrow 2LiOH + H_2$, shows the stoichiometric relationship between reactants and products. For every 2 moles of lithium, 1 mole of hydrogen gas is formed. This 2:1 molar ratio is key for the conversion.
However, the reaction did not proceed with 100% efficiency. Instead, the reaction yield was 85%, meaning the amount of hydrogen gas actually collected (0.30 mol) is less than what would be predicted if all lithium reacted perfectly. This necessitates calculating the theoretical yield, the amount of hydrogen that would be produced with 100% efficiency, using the formula:
$$
\text{Theoretical yield} = \frac{\text{Actual yield}}{\text{Percent yield}}
$$
Plugging in the numbers: $0.30 \div 0.85 \approx 0.3529$ mol H₂. This is the amount of hydrogen that would be formed if the reaction had gone to completion with no loss.
Next, using the stoichiometric ratio, we find the amount of lithium required to produce that theoretical amount of H₂. Since 2 mol of Li make 1 mol of H₂, for 0.3529 mol of H₂ we need $2 \times 0.3529 = 0.7058$ mol of Li.
After rounding appropriately, the final answer is 0.71 mol of lithium reacted. This approach ensures accurate, chemistry-based reasoning to solve real-world laboratory problems
.
