Question

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

Short answer

The theoretical yield of Li₃N is 20.58 g, and the percent yield is 28.62%.

Step-by-step solution

Convert Mass to Moles

To find the theoretical yield, first convert the masses of the reactants to moles. - The molar mass of Lithium (Li) = 6.94 g/mol.- The molar mass of Nitrogen gas (N₂) = 28.02 g/mol.Calculate moles of Li: \[ \text{Moles of Li} = \frac{12.3 \text{ g}}{6.94 \text{ g/mol}} \approx 1.773 \text{ moles} \]Calculate moles of N₂:\[ \text{Moles of N}_2 = \frac{33.6 \text{ g}}{28.02 \text{ g/mol}} \approx 1.199 \text{ moles} \]

Determine the Limiting Reactant

The balanced equation is:\[ 6 \text{ Li} + \text{ N}_2 \to 2 \text{ Li}_3\text{N} \]The molar ratio of Li to N₂ is 6:1. Calculate the amount of N₂ required to react with 1.773 moles of Li:\[ \text{Required moles of N}_2 = \frac{1.773 \text{ moles Li}}{6} \approx 0.296 \text{ moles N}_2 \]Since 0.296 moles of N₂ are needed and 1.199 moles are available, Li is the limiting reactant.

Calculate Theoretical Yield of Li₃N

The balanced equation also demonstrates that 6 moles of Li produce 2 moles of Li₃N.Calculate the moles of Li₃N produced from 1.773 moles of Li:\[ \text{Moles of Li}_3\text{N} = \frac{1.773 \text{ moles Li} \times 2}{6} \approx 0.591 \text{ moles Li}_3\text{N} \]The molar mass of Li₃N = 34.83 g/mol.Calculate the theoretical yield in grams:\[ \text{Theoretical yield} = 0.591 \text{ moles} \times 34.83 \text{ g/mol} \approx 20.58 \text{ g} \]

Calculate Percent Yield

The actual yield of Li₃N is given as 5.89 g.Percent yield is calculated using the formula:\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\%\]\[ \text{Percent yield} = \left( \frac{5.89 \text{ g}}{20.58 \text{ g}} \right) \times 100\% \approx 28.62\% \]

Key concepts

Limiting Reactant

In many chemical reactions, the limiting reactant plays a crucial role in determining how much product can be formed. This reactant is the substance that is entirely consumed first, effectively stopping the reaction, because there is no more of this substance available to react. It's like trying to make a sandwich but running out of bread - the bread limits how many sandwiches you can make.

To figure out which reactant is the limiting one, you must compare the mole ratio of the available reactants to the mole ratio in the balanced chemical equation. Take the given exercise: we have 1.773 moles of lithium (Li) and 1.199 moles of nitrogen gas (N extsubscript{2}). The chemical equation tells us that 6 moles of Li react with 1 mole of N extsubscript{2}.

• Calculate how much N extsubscript{2} is needed for the 1.773 moles of Li: \[\text{Required moles of N}_2 = \frac{1.773 \text{ moles Li}}{6} \approx 0.296 \text{ moles N}_2\]
• Since 0.296 moles of N extsubscript{2} are required and we actually have 1.199 moles, it is clear that all the Li will be used up first, making Li the limiting reactant.

Understanding the concept of the limiting reactant is foundational in predicting the theoretical amount of product a reaction can produce.

Percent Yield Calculation

Percent yield is a valuable concept in chemistry as it gives insight into the efficiency of a reaction. It compares the actual yield, what you really get from your reaction, with the theoretical yield, or what you could ideally produce under perfect conditions.

In the given problem, we already calculated the theoretical yield of lithium nitride (Li extsubscript{3}N) to be approximately 20.58 grams. However, the actual yield was found to be 5.89 grams. The formula for calculating percent yield is:\[\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\%\]Using the data given:

• \[\text{Percent yield} = \left( \frac{5.89 \text{ g}}{20.58 \text{ g}} \right) \times 100\% \approx 28.62\%\]

This calculated percent yield of 28.62% tells us that the reaction was not very efficient, and various factors such as loss of product during handling or side reactions could have contributed to the lower yield. Investigating ways to improve percent yield is often a key task in experimental chemistry.

Moles to Grams Conversion

One fundamental skill in chemistry, especially when dealing with reactions, is converting between moles and grams. This conversion is essential as laboratory scales measure mass in grams, but chemical equations work with moles.

To convert from moles to grams, you simply need to multiply the number of moles by the molar mass of the substance, which provides you with grams. This is because 1 mole of any substance contains exactly Avogadro's number (approximately 6.022 x 10\(^{23}\)) of entities and weighs its molar mass.

Here's a simple example from our exercise involving lithium nitride (Li extsubscript{3}N). Say you've calculated you've got 0.591 moles of it. The molar mass of Li extsubscript{3}N is given as 34.83 g/mol. The conversion to grams is straightforward:

• \[\text{Grams of Li}_3\text{N} = 0.591 \text{ moles} \times 34.83 \text{ g/mol} \approx 20.58 \text{ g}\]

Grasping the concept of moles to grams conversion is vital, as it bridges the gap between the theoretical world of chemical equations and practical laboratory measurements.