Stoichiometry: $Mg_3N_2$ Production Calculation

Hey guys! Ever find yourself staring at a chemical equation and wondering just how much product you can actually make? It's a common challenge in chemistry, and today we're diving deep into a stoichiometry problem to figure out exactly that. We'll be tackling a specific reaction and calculating the mass of magnesium nitride ($Mg_3N_2$) produced. So, grab your calculators, and let's get started!

Unpacking the Reaction: $3MgCO_3 + Ba_3N_2

ightarrow Mg_3N_2 + 3BaCO_3$

Before we jump into the calculations, let's break down the chemical equation we're working with:

$3MgCO_3 + Ba_3N_2 ightarrow Mg_3N_2 + 3BaCO_3$

This equation tells us that three molecules of magnesium carbonate ($MgCO_3$) react with one molecule of barium nitride ($Ba_3N_2$) to produce one molecule of magnesium nitride ($Mg_3N_2$) and three molecules of barium carbonate ($BaCO_3$). This balanced equation is our roadmap, giving us the crucial mole ratios needed for our calculations. Stoichiometry, at its core, is all about understanding these relationships and using them to predict the amounts of reactants and products involved in a chemical reaction. The coefficients in front of each chemical formula are key; they represent the number of moles of each substance participating in the reaction. For example, the '3' in front of $MgCO_3$ indicates that three moles of magnesium carbonate react for every one mole of barium nitride. These mole ratios are the foundation upon which we build our calculations. Without a balanced equation, we'd be lost in the stoichiometric wilderness! In this specific reaction, magnesium carbonate and barium nitride are our reactants – the substances we're starting with. Magnesium nitride and barium carbonate are our products – the substances that are formed as a result of the reaction. Our goal is to determine how much magnesium nitride we can produce given specific amounts of our reactants. To do this, we'll need to convert the masses of our reactants into moles, use the mole ratios from the balanced equation to determine the limiting reactant, and then calculate the theoretical yield of magnesium nitride. So, as you can see, there are several steps involved, but each one is crucial to arriving at the correct answer. We'll take it step-by-step, making sure each concept is clear before moving on. Remember, stoichiometry isn't about memorizing formulas; it's about understanding the fundamental relationships between chemical substances in a reaction. So, let's keep that in mind as we proceed through this problem. It's like following a recipe, but instead of baking a cake, we're 'baking' a chemical product! The balanced equation is our recipe, and the mole ratios are our ingredient proportions. If we follow the recipe correctly, we'll end up with the desired amount of product. And that's the beauty of stoichiometry – it allows us to predict and control the outcomes of chemical reactions. Now, with our balanced equation in hand and a solid understanding of its importance, let's move on to the next step: converting the given masses of reactants into moles. This is a critical step because stoichiometry works with moles, not grams. So, buckle up, and let's dive into the world of molar masses and mole conversions! Remember, we're on a mission to figure out how much magnesium nitride we can produce, and we're well on our way. So, stay focused, and let's keep those calculations coming!

The Given Information: 1.2 g of $MgCO_3$ and 2.5 g of $Ba_3N_2$

Okay, so we know that 1.2 grams of magnesium carbonate ($MgCO_3$) are reacting with 2.5 grams of barium nitride ($Ba_3N_2$). These are our starting amounts, but grams aren't directly useful in stoichiometry. We need to convert these masses into moles – the chemist's favorite unit for measuring amounts of substances. Think of it like this: grams are like measuring ingredients by weight, while moles are like measuring them by volume. In a recipe, you need to know the volume of each ingredient to get the proportions right. Similarly, in a chemical reaction, we need to know the number of moles of each reactant to figure out how much product we can make. The key to converting grams to moles is the molar mass. The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). You can calculate the molar mass by adding up the atomic masses of all the atoms in the chemical formula. Atomic masses are found on the periodic table. For example, to find the molar mass of magnesium carbonate ($MgCO_3$), we need the atomic masses of magnesium (Mg), carbon (C), and oxygen (O). Magnesium has an atomic mass of approximately 24.31 g/mol, carbon is about 12.01 g/mol, and oxygen is roughly 16.00 g/mol. Since there are three oxygen atoms in $MgCO_3$, we multiply oxygen's atomic mass by 3. So, the molar mass of $MgCO_3$ is (24.31 g/mol) + (12.01 g/mol) + 3*(16.00 g/mol) = 84.32 g/mol. Similarly, we can calculate the molar mass of barium nitride ($Ba_3N_2$). Barium (Ba) has an atomic mass of approximately 137.33 g/mol, and nitrogen (N) is about 14.01 g/mol. Since there are three barium atoms and two nitrogen atoms, the molar mass of $Ba_3N_2$ is 3*(137.33 g/mol) + 2*(14.01 g/mol) = 439.99 g/mol. Now that we have the molar masses, we can finally convert the given masses into moles. To do this, we simply divide the mass of each substance by its molar mass. For magnesium carbonate, we have 1.2 g / 84.32 g/mol = 0.0142 moles (approximately). For barium nitride, we have 2.5 g / 439.99 g/mol = 0.0057 moles (approximately). So, we now know that we're starting with about 0.0142 moles of $MgCO_3$ and 0.0057 moles of $Ba_3N_2$. This is a crucial step forward because we can now use these mole amounts and the mole ratios from the balanced equation to determine which reactant is the limiting reactant. The limiting reactant is the reactant that gets used up first, and it's the one that determines the maximum amount of product that can be formed. Identifying the limiting reactant is like figuring out which ingredient you'll run out of first when baking a cake – you can only make as much cake as you have of that ingredient! So, let's move on to the next step and find out which of our reactants is the limiting one. We're getting closer to calculating the amount of magnesium nitride produced, so let's keep the momentum going!

Calculating Moles: Converting Grams to Moles

Alright, guys, time to crunch some numbers! We've got our starting masses of the reactants, and we know we need to convert them to moles. This is a fundamental step in stoichiometry because chemical reactions happen on a mole-to-mole basis, not a gram-to-gram basis. Think of it like following a recipe – you don't just throw in random amounts of ingredients; you need specific ratios to get the desired outcome. Similarly, in a chemical reaction, the mole ratios between reactants and products dictate how much product can be formed. So, how do we convert grams to moles? The magic tool is the molar mass. As we discussed earlier, the molar mass is the mass of one mole of a substance, and it's expressed in grams per mole (g/mol). We calculate the molar mass by adding up the atomic masses of all the atoms in the chemical formula, which we can find on the periodic table. Let's start with magnesium carbonate ($MgCO_3$). The molar mass of magnesium (Mg) is approximately 24.31 g/mol, carbon (C) is about 12.01 g/mol, and oxygen (O) is roughly 16.00 g/mol. Since there are three oxygen atoms in $MgCO_3$, we need to multiply oxygen's atomic mass by 3. So, the molar mass of $MgCO_3$ is: (24.31 g/mol) + (12.01 g/mol) + 3*(16.00 g/mol) = 84.32 g/mol. Now, we can use this molar mass to convert the given mass of $MgCO_3$ (1.2 g) into moles: Moles of $MgCO_3$ = (Mass of $MgCO_3$) / (Molar mass of $MgCO_3$) = 1.2 g / 84.32 g/mol ≈ 0.0142 moles. Next, let's tackle barium nitride ($Ba_3N_2$). The molar mass of barium (Ba) is approximately 137.33 g/mol, and nitrogen (N) is about 14.01 g/mol. Since there are three barium atoms and two nitrogen atoms in $Ba_3N_2$, the molar mass of $Ba_3N_2$ is: 3*(137.33 g/mol) + 2*(14.01 g/mol) = 439.99 g/mol. Now, we can convert the given mass of $Ba_3N_2$ (2.5 g) into moles: Moles of $Ba_3N_2$ = (Mass of $Ba_3N_2$) / (Molar mass of $Ba_3N_2$) = 2.5 g / 439.99 g/mol ≈ 0.0057 moles. So, we've successfully converted the masses of our reactants into moles! We now know that we have approximately 0.0142 moles of $MgCO_3$ and 0.0057 moles of $Ba_3N_2$. This is a crucial milestone because we can now use these mole amounts and the balanced chemical equation to determine the limiting reactant. Remember, the limiting reactant is the reactant that gets used up first, and it determines the maximum amount of product that can be formed. Identifying the limiting reactant is like figuring out which ingredient you'll run out of first when baking a cake – you can only make as much cake as you have of that ingredient! So, let's move on to the next step and figure out which of our reactants is the limiting one. We're getting closer to calculating the amount of magnesium nitride produced, so let's keep the calculations rolling!

Identifying the Limiting Reactant: Which Runs Out First?

Okay, folks, this is a crucial step! We need to figure out which reactant is going to limit the amount of product we can make. This is where the concept of the limiting reactant comes into play. The limiting reactant is like the shortest ingredient in a recipe – you can only make as much of the final dish as you have of that ingredient, even if you have plenty of everything else. In our reaction, the limiting reactant is the one that gets completely used up first, thus stopping the reaction and determining the maximum amount of product that can be formed. So, how do we identify the limiting reactant? We'll use the mole ratios from our balanced equation: $3MgCO_3 + Ba_3N_2 ightarrow Mg_3N_2 + 3BaCO_3$. The balanced equation tells us that 3 moles of $MgCO_3$ react with 1 mole of $Ba_3N_2$. This is our key ratio. We have 0.0142 moles of $MgCO_3$ and 0.0057 moles of $Ba_3N_2$. To figure out which is the limiting reactant, we can compare the mole ratio of the reactants we have to the mole ratio from the balanced equation. One way to do this is to calculate how much of one reactant is needed to react completely with the other reactant. Let's start by figuring out how many moles of $MgCO_3$ we need to react completely with 0.0057 moles of $Ba_3N_2$. Using the mole ratio from the balanced equation (3 moles $MgCO_3$ / 1 mole $Ba_3N_2$), we can calculate: Moles of $MgCO_3$ needed = 0.0057 moles $Ba_3N_2$ * (3 moles $MgCO_3$ / 1 mole $Ba_3N_2$) = 0.0171 moles $MgCO_3$. This means we need 0.0171 moles of $MgCO_3$ to react completely with the 0.0057 moles of $Ba_3N_2$ we have. But we only have 0.0142 moles of $MgCO_3$. Since we don't have enough $MgCO_3$ to react with all the $Ba_3N_2$, $MgCO_3$ is our limiting reactant! Another way to think about it is to calculate how much $Ba_3N_2$ we need to react completely with 0.0142 moles of $MgCO_3$: Moles of $Ba_3N_2$ needed = 0.0142 moles $MgCO_3$ * (1 mole $Ba_3N_2$ / 3 moles $MgCO_3$) = 0.0047 moles $Ba_3N_2$. We have 0.0057 moles of $Ba_3N_2$, which is more than the 0.0047 moles needed. This confirms that $MgCO_3$ is the limiting reactant because we have less of it than we need to react with all the $Ba_3N_2$. So, we've identified our limiting reactant: $MgCO_3$! This is a huge step because it tells us that the amount of $MgCO_3$ we have will determine the maximum amount of $Mg_3N_2$ we can produce. We can now use the amount of the limiting reactant and the mole ratio from the balanced equation to calculate the theoretical yield of our product. The theoretical yield is the maximum amount of product that can be formed from a given amount of reactants, assuming the reaction goes to completion. In reality, we might not get the full theoretical yield due to various factors, but it's a useful benchmark for our calculations. So, let's move on to the next step and calculate the theoretical yield of $Mg_3N_2$. We're on the home stretch now, guys! Let's keep those calculations coming and get to our final answer!

Calculating Theoretical Yield of $Mg_3N_2$

Alright, we've identified our limiting reactant as magnesium carbonate ($MgCO_3$). This means the amount of $MgCO_3$ we have will dictate the maximum amount of magnesium nitride ($Mg_3N_2$) we can produce. Now, let's calculate the theoretical yield – the maximum amount of product we can get if everything goes perfectly. To do this, we'll use the mole ratio between $MgCO_3$ and $Mg_3N_2$ from our balanced equation: $3MgCO_3 + Ba_3N_2 ightarrow Mg_3N_2 + 3BaCO_3$. The equation tells us that 3 moles of $MgCO_3$ produce 1 mole of $Mg_3N_2$. This is the key ratio we need for our calculation. We know we have 0.0142 moles of $MgCO_3$ (our limiting reactant). Using the mole ratio, we can calculate the moles of $Mg_3N_2$ produced: Moles of $Mg_3N_2$ = 0.0142 moles $MgCO_3$ * (1 mole $Mg_3N_2$ / 3 moles $MgCO_3$) = 0.00473 moles $Mg_3N_2$ (approximately). So, theoretically, we can produce 0.00473 moles of $Mg_3N_2$. But the question asks for the answer in grams, not moles. So, we need to convert moles back to grams. To do this, we'll use the molar mass of $Mg_3N_2$. We need the atomic masses of magnesium (Mg) and nitrogen (N). Magnesium is approximately 24.31 g/mol, and nitrogen is about 14.01 g/mol. In $Mg_3N_2$, there are three magnesium atoms and two nitrogen atoms. So, the molar mass of $Mg_3N_2$ is: 3*(24.31 g/mol) + 2*(14.01 g/mol) = 100.95 g/mol. Now we can convert moles of $Mg_3N_2$ to grams: Grams of $Mg_3N_2$ = (Moles of $Mg_3N_2$) * (Molar mass of $Mg_3N_2$) = 0.00473 moles * 100.95 g/mol = 0.477 grams (approximately). So, the theoretical yield of $Mg_3N_2$ is approximately 0.477 grams. This is the maximum amount of $Mg_3N_2$ we can produce from 1.2 g of $MgCO_3$ and 2.5 g of $Ba_3N_2$, assuming the reaction goes to completion. It's important to remember that this is a theoretical yield. In a real-world experiment, we might not get this exact amount due to factors like incomplete reactions or loss of product during purification. But it gives us a good estimate of what to expect. We did it, guys! We've successfully calculated the theoretical yield of $Mg_3N_2$ using stoichiometry. We converted grams to moles, identified the limiting reactant, used mole ratios from the balanced equation, and converted back to grams. This is a classic stoichiometry problem, and you've now mastered the steps to solve it. So, next time you encounter a similar problem, remember the process and tackle it with confidence! You've got this!

Final Answer: Approximately 0.477 grams of $Mg_3N_2$ Will Be Produced

Alright, guys, let's recap! We've journeyed through the world of stoichiometry, tackled a balanced chemical equation, and calculated the theoretical yield of magnesium nitride ($Mg_3N_2$). We started with the reaction: $3MgCO_3 + Ba_3N_2 ightarrow Mg_3N_2 + 3BaCO_3$. We were given 1.2 grams of magnesium carbonate ($MgCO_3$) and 2.5 grams of barium nitride ($Ba_3N_2$) and asked to find out how many grams of $Mg_3N_2$ would be produced. The first crucial step was converting the given masses into moles. We calculated the molar masses of $MgCO_3$ (84.32 g/mol) and $Ba_3N_2$ (439.99 g/mol) and used them to convert grams to moles: 1. 2 g $MgCO_3$ ≈ 0.0142 moles 2. 5 g $Ba_3N_2$ ≈ 0.0057 moles. Next, we identified the limiting reactant. By comparing the mole ratio of the reactants to the stoichiometry of the balanced equation, we determined that $MgCO_3$ was the limiting reactant. This means that the amount of $MgCO_3$ we have dictates the maximum amount of $Mg_3N_2$ we can produce. With the limiting reactant in hand, we calculated the theoretical yield of $Mg_3N_2$. Using the mole ratio from the balanced equation (3 moles $MgCO_3$ : 1 mole $Mg_3N_2$), we found that 0.0142 moles of $MgCO_3$ could produce approximately 0.00473 moles of $Mg_3N_2$. Finally, we converted moles of $Mg_3N_2$ back to grams using its molar mass (100.95 g/mol): 0. 00473 moles $Mg_3N_2$ ≈ 0.477 grams $Mg_3N_2$. So, drumroll please… Our final answer is that approximately 0.477 grams of $Mg_3N_2$ will be produced when 1.2 g of $MgCO_3$ reacts with 2.5 g of $Ba_3N_2$. Woohoo! We did it! Stoichiometry can seem daunting at first, but by breaking it down into smaller steps – converting to moles, identifying the limiting reactant, using mole ratios, and converting back to grams – we can conquer these types of problems with confidence. Remember, chemistry is like a puzzle, and stoichiometry is a key piece to solving it. So, keep practicing, keep exploring, and keep those calculations coming! You've got the tools, now go out there and rock those chemistry problems! And that’s a wrap, guys! I hope this breakdown helped you understand how to tackle stoichiometry problems. Keep shining, keep learning, and I’ll catch you in the next one!