Calculating Li3N Needed For 24g LiOH: A Chemistry Problem

Oct 27, 2025 by Admin 58 views

Hey guys! Let's dive into a fun chemistry problem today. We're going to figure out how much lithium nitride ( $Li_3N$ ) we need to react with water to produce 24 grams of lithium hydroxide (LiOH). This is a classic stoichiometry problem, and we'll break it down step by step so it's super easy to follow. So, grab your calculators and let's get started!

Understanding the Reaction

First, let's take a close look at the balanced chemical equation:

$Li_3N(s) + 3H_2O(l) ightarrow NH_3(g) + 3LiOH(l)$

This equation tells us that one mole of solid lithium nitride ( $Li_3N$ ) reacts with three moles of liquid water ( $H_2O$ ) to produce one mole of gaseous ammonia ( $NH_3$ ) and three moles of lithium hydroxide ( $LiOH$ ). This balanced equation is the key to solving our problem because it gives us the mole ratios between the reactants and products.

Why is the Balanced Equation So Important?

The balanced equation is crucial because it adheres to the law of conservation of mass. This law states that matter cannot be created or destroyed in a chemical reaction. By balancing the equation, we ensure that the number of atoms of each element is the same on both sides of the equation. This allows us to accurately predict the amount of reactants needed and products formed in a chemical reaction. Without a balanced equation, our calculations would be way off, and we wouldn't get the correct answer. It's like trying to bake a cake without measuring the ingredients – you might end up with something completely different from what you intended!

Molar Masses: Your Best Friends in Stoichiometry

To convert between grams and moles, we need to use molar masses. The molar mass of a compound is the mass of one mole of that compound, usually expressed in grams per mole (g/mol). We can calculate the molar mass by adding up the atomic masses of all the atoms in the compound. You can find the atomic masses on the periodic table.

Lithium Nitride ( $Li_3N$ ): (3 * Li) + (1 * N) = (3 * 6.94 g/mol) + (1 * 14.01 g/mol) = 20.82 g/mol + 14.01 g/mol = 34.83 g/mol
Lithium Hydroxide (LiOH): (1 * Li) + (1 * O) + (1 * H) = (1 * 6.94 g/mol) + (1 * 16.00 g/mol) + (1 * 1.01 g/mol) = 6.94 g/mol + 16.00 g/mol + 1.01 g/mol = 23.95 g/mol

These molar masses are the conversion factors we'll use to switch between grams and moles. Think of them as the bridge between the mass world (grams) and the mole world, where chemical reactions happen!

Step-by-Step Calculation

Now that we have the balanced equation and the molar masses, we can tackle the problem. Here’s the plan:

Convert grams of LiOH to moles of LiOH.
Use the mole ratio from the balanced equation to find moles of $Li_3N$ needed.
Convert moles of $Li_3N$ to grams of $Li_3N$ .

Step 1: Grams of LiOH to Moles of LiOH

We want to make 24 grams of LiOH. To convert this to moles, we'll use the molar mass of LiOH (23.95 g/mol):

Moles of LiOH = (Grams of LiOH) / (Molar mass of LiOH) Moles of LiOH = (24 g) / (23.95 g/mol) = 1.002 moles

So, we need to produce approximately 1.002 moles of LiOH. This is our first major milestone! We've successfully converted the mass of LiOH into moles, which is the language of chemical reactions.

Step 2: Mole Ratio to Find Moles of $Li_3N$

Now, we'll use the balanced chemical equation to find the mole ratio between $Li_3N$ and LiOH. From the equation:

$Li_3N(s) + 3H_2O(l) ightarrow NH_3(g) + 3LiOH(l)$

We see that 1 mole of $Li_3N$ produces 3 moles of LiOH. This gives us the mole ratio:

(1 mole $Li_3N$ ) / (3 moles LiOH)

We can use this ratio to find out how many moles of $Li_3N$ we need to react to produce 1.002 moles of LiOH:

Moles of $Li_3N$ = (Moles of LiOH) * (Mole ratio of $Li_3N$ to LiOH) Moles of $Li_3N$ = (1.002 moles LiOH) * (1 mole $Li_3N$ / 3 moles LiOH) = 0.334 moles $Li_3N$

Therefore, we need 0.334 moles of $Li_3N$ to react with excess water. We're getting closer to our final answer!

Step 3: Moles of $Li_3N$ to Grams of $Li_3N$

Finally, we need to convert moles of $Li_3N$ to grams. We'll use the molar mass of $Li_3N$ (34.83 g/mol):

Grams of $Li_3N$ = (Moles of $Li_3N$ ) * (Molar mass of $Li_3N$ ) Grams of $Li_3N$ = (0.334 moles) * (34.83 g/mol) = 11.64 g

The Grand Finale: The Answer!

So, to make 24 grams of LiOH, you must react approximately 11.64 grams of $Li_3N$ with excess water. Awesome, right? We've successfully navigated through the stoichiometry and arrived at our answer. Remember, chemistry is all about following the steps and understanding the relationships between molecules.

Key Takeaways

Let's recap the key concepts we've used in this problem:

Balanced Chemical Equations: These are essential for determining mole ratios between reactants and products.
Molar Mass: This is the conversion factor between grams and moles.
Mole Ratios: These allow us to calculate the amount of reactants needed or products formed in a reaction.

Practice Makes Perfect!

If you found this helpful, try practicing more stoichiometry problems. The more you practice, the better you'll become at these calculations. You can even try changing the amount of LiOH you want to produce and recalculating the amount of $Li_3N$ needed. Experiment and have fun with chemistry!

Stoichiometry in the Real World

You might be wondering, why is this stuff important? Well, stoichiometry is used in many real-world applications, such as:

Pharmaceuticals: Calculating the correct amount of reactants to synthesize drugs.
Manufacturing: Ensuring the correct proportions of ingredients in chemical products.
Environmental Science: Determining the amount of pollutants produced in a reaction.
Cooking: Yes, even cooking! Recipes are essentially stoichiometric equations, telling you the ratios of ingredients needed.

So, understanding stoichiometry is not just about passing a chemistry test; it's a valuable skill that can be applied in many areas of life.

Conclusion

We've successfully calculated the amount of $Li_3N$ needed to produce 24 grams of LiOH. Remember, chemistry problems can seem daunting at first, but by breaking them down into smaller steps, we can solve even the trickiest ones. Keep practicing, stay curious, and most importantly, have fun with chemistry! You guys rock! If you have any questions or want to try another problem, just let me know!