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 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:
To determine how many moles of lithium (Li) reacted, let’s go through the steps systematically.
Step 1: Understanding the Reaction
The balanced chemical equation given is:
2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)2Li (s) + 2H_2O (l) \rightarrow 2LiOH (aq) + H_2 (g)
From the equation, 2 moles of Li produce 1 mole of H₂ gas.
Step 2: Actual Yield vs. Theoretical Yield
The problem states that 0.30 moles of H₂ gas are isolated in the lab at an 85% yield. This means the actual yield is 85% of the theoretical amount.
Using the formula:
Actual yield=Percent yield×Theoretical yield\text{Actual yield} = \text{Percent yield} \times \text{Theoretical yield}
Rearranging to find the theoretical yield:
Theoretical yield=Actual yieldPercent yield\text{Theoretical yield} = \frac{\text{Actual yield}}{\text{Percent yield}}
Substituting values:
Theoretical yield=0.30 mol0.85=0.35 mol H2\text{Theoretical yield} = \frac{0.30 \text{ mol}}{0.85} = 0.35 \text{ mol H}_2
Step 3: Finding the Moles of Lithium Reacted
From the balanced equation:
- 2 moles of Li produce 1 mole of H₂.
- Therefore, to produce 0.35 moles of H₂, the amount of lithium needed is:
Moles of Li=0.35 mol H2×2 mol Li1 mol H2\text{Moles of Li} = 0.35 \text{ mol H}_2 \times \frac{2 \text{ mol Li}}{1 \text{ mol H}_2}
=0.70 mol Li= 0.70 \text{ mol Li}
Explanation:
The key idea in solving this problem is understanding stoichiometric relationships in the balanced equation. Since 2 moles of lithium produce 1 mole of hydrogen gas, we used that ratio to determine the theoretical amount of hydrogen gas needed before adjusting for the 85% yield. Dividing the actual yield by 0.85 allowed us to find the theoretical yield, and from there, we calculated the lithium required using molar ratios. The answer is 0.70 moles of lithium reacted to form the hydrogen gas collected in the experiment.
