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When heated, lithium reacts with nitrogen to form lithium nitride: $$ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s) $$ What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) are heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2} ?\) If the actual yield of \(\mathrm{Li}_{3} \mathrm{~N}\) is \(5.89 \mathrm{~g},\) what is the percent yield of the reaction?

Short Answer

Expert verified
The theoretical yield is 20.590 g, and the percent yield is 28.6%.

Step by step solution

01

Find the molar mass of reactants and product

Calculate the molar mass of lithium (\(\mathrm{Li}\)), nitrogen (\(\mathrm{N}_2\)), and lithium nitride (\(\mathrm{Li}_3\mathrm{N}\)). - Molar mass of \(\mathrm{Li} = 6.94 \, \mathrm{g/mol}\)- Molar mass of \(\mathrm{N}_2 = 28.02 \, \mathrm{g/mol}\)- Molar mass of \(\mathrm{Li}_3\mathrm{N} = 3 \times 6.94 + 14.01 = 34.83 \, \mathrm{g/mol}\)
02

Determine the limiting reactant

Convert grams of each reactant to moles and compare using the balanced equation.- Moles of \(\mathrm{Li} = \frac{12.3\, \mathrm{g}}{6.94\, \mathrm{g/mol}} = 1.772\, \text{mol}\)- Moles of \(\mathrm{N}_2 = \frac{33.6\, \mathrm{g}}{28.02\, \mathrm{g/mol}} = 1.199\, \text{mol}\)According to the equation, 6 moles of \(\mathrm{Li}\) react with 1 mole of \(\mathrm{N}_2\). Therefore,- Required moles of \(\mathrm{Li}\) for 1.199 mol \(\mathrm{N}_2\): \(6 \times 1.199 = 7.194\, \text{mol}\)Since 1.772 mol < 7.194 mol, \(\mathrm{Li}\) is the limiting reactant.
03

Calculate the theoretical yield

Use the moles of the limiting reactant (\(\mathrm{Li}\)) to find the moles of \(\mathrm{Li}_3\mathrm{N}\) produced. From the balanced equation, 6 moles of \(\mathrm{Li}\) produce 2 moles of \(\mathrm{Li}_3\mathrm{N}\).- Moles of \(\mathrm{Li}_3\mathrm{N}\) from \(\mathrm{Li}\): \(\frac{1.772}{6} \times 2 = 0.59133 \, \text{mol}\)Convert moles of \(\mathrm{Li}_3\mathrm{N}\) to grams using its molar mass:- Theoretical yield: \(0.59133 \, \text{mol} \times 34.83 \, \mathrm{g/mol} = 20.590 \, \mathrm{g}\)
04

Calculate the percent yield

Percent yield is calculated using the actual yield and the theoretical yield.\[\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% \]- Percent yield: \( \left( \frac{5.89 \, \mathrm{g}}{20.590 \, \mathrm{g}} \right) \times 100\% = 28.6\% \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Theoretical Yield
When performing chemical reactions, the theoretical yield is an essential concept that represents the maximum amount of product that could theoretically be obtained from given quantities of reactants.
To calculate the theoretical yield, one must first understand the balanced chemical equation, which shows the ratio of reactants to products. Using stoichiometry, where the moles of the limiting reactant are converted to moles of the product, we can predict the maximum amount of product formed.
In the original exercise, we calculated the theoretical yield of lithium nitride ( Li_3N ) by first identifying lithium ( Li ) as the limiting reactant. By using the stoichiometry from the balanced equation: 6 moles of lithium produce 2 moles of lithium nitride. Here, we only had 1.772 moles of lithium, which gave us 0.59133 moles of lithium nitride.
Lastly, we converted the moles of lithium nitride to grams using its molar mass, arriving at a theoretical yield of 20.590 grams.
Limiting Reactant
The limiting reactant is the reactant in a chemical reaction that is completely consumed first, thus determining the amount of product formed. It is crucial for calculating the theoretical yield, ensuring no reactants are assumed to produce more product than is possible.
To identify the limiting reactant, we convert the mass of each reactant to moles and compare their ratios based on the balanced equation. In the exercise, we found that lithium ( Li ) was the limiting reactant.
This was determined by calculating: 6 moles of Li were needed to react with 1 mole of N_2 . Given 1.199 moles of N_2 were available, the required lithium was 7.194 moles, exceeding what was actually present (1.772 moles). Therefore, lithium limited the reaction.
Percent Yield
Percent yield is an important concept used to evaluate the efficiency of a chemical reaction. It tells us how much of the theoretical yield was actually achieved when the reaction was performed.
To calculate percent yield, the actual yield obtained from the experiment is divided by the theoretical yield, and the result is multiplied by 100% for a percentage.
In the given problem, the actual yield of lithium nitride (Li_3N) was 5.89 grams, while the theoretical yield was 20.590 grams. Thus, the percent yield is calculated as follows: \[ \text{Percent yield} = \left( \frac{5.89 \, \mathrm{g}}{20.590 \, \mathrm{g}} \right) \times 100\% \approx 28.6\% \]
This means the reaction produced only 28.6% of what was theoretically possible, indicating either incomplete reactions or potential losses during the procedure.

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