Nitrogen Needed For 265g Calcium Cyanamide
Hey there, chemistry enthusiasts! Ever wondered about the nitty-gritty of chemical reactions and how much of one substance you need to get a specific amount of another? Today, we're diving deep into a fascinating reaction to figure out just that. We're going to break down the reaction between calcium carbide ($CaC_2$) and nitrogen ($N_2$) to produce calcium cyanamide ($CaCN_2$) and carbon ($C$). Specifically, we'll calculate how many grams of nitrogen gas ($N_2$) are required to produce 265 grams of calcium cyanamide ($CaCN_2$). Buckle up, because we're about to get into the wonderful world of stoichiometry!
The Chemical Equation: A Recipe for Reactions
First things first, let's take a look at the balanced chemical equation. This equation is like a recipe for a chemical reaction; it tells us exactly what we need and what we'll get. The equation we're working with is:
This equation tells us that one mole of calcium carbide ($CaC_2$) reacts with one mole of nitrogen gas ($N_2$) to produce one mole of calcium cyanamide ($CaCN_2$) and one mole of carbon ($C$). It's a 1:1:1:1 molar ratio, which makes our calculations a bit simpler. Now, why is this balanced equation so important? Well, it ensures that the number of atoms for each element is the same on both sides of the equation, adhering to the law of conservation of mass. This law is a cornerstone of chemistry, stating that matter cannot be created or destroyed in a chemical reaction. So, what goes in must come out, just in a different form.
Molar Mass: The Bridge Between Grams and Moles
Now, before we can dive into the calculations, we need to talk about 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). Think of it as the weight of a specific number of molecules (Avogadro's number, to be exact). To solve our problem, we need the molar masses of nitrogen ($N_2$) and calcium cyanamide ($CaCN_2$). The molar mass of nitrogen ($N_2$) is approximately 28.02 g/mol (since nitrogen has an atomic mass of about 14.01 g/mol, and we have two nitrogen atoms). For calcium cyanamide ($CaCN_2$), we need to add up the atomic masses of one calcium atom (Ca, approximately 40.08 g/mol), one carbon atom (C, approximately 12.01 g/mol), and two nitrogen atoms (N, approximately 14.01 g/mol each). So, the molar mass of $CaCN_2$ is approximately:
- 08 + 12.01 + (2 * 14.01) = 80.11 g/mol
Understanding molar mass is absolutely crucial because it acts as the bridge between the grams we're given in the problem (265 grams of $CaCN_2$) and the moles we need to work with in the balanced equation. Moles are the currency of chemical reactions; they allow us to directly compare the amounts of different substances involved. The molar mass concept is not just a number; it's a fundamental property that connects the macroscopic world (grams) to the microscopic world (moles and molecules). So, when you're tackling stoichiometry problems, always start by calculating those molar masses!
Stoichiometry: The Art of Calculation
Okay, now for the fun part: the stoichiometry! Stoichiometry is basically the math of chemical reactions. It allows us to predict the amounts of reactants and products involved in a reaction. In our case, we want to find out how many grams of $N_2$ are needed to produce 265 grams of $CaCN_2$.
Step 1: Convert Grams of $CaCN_2$ to Moles
We know we want to produce 265 grams of $CaCN_2$, and we know the molar mass of $CaCN_2$ is 80.11 g/mol. To convert grams to moles, we use the following formula:
Moles = Grams / Molar Mass
So, the moles of $CaCN_2$ produced are:
Moles of $CaCN_2$ = 265 g / 80.11 g/mol ≈ 3.308 moles
Step 2: Use the Molar Ratio from the Balanced Equation
Remember our balanced equation? It tells us that the molar ratio of $N_2$ to $CaCN_2$ is 1:1. This means that for every one mole of $CaCN_2$ produced, we need one mole of $N_2$. Since we're producing 3.308 moles of $CaCN_2$, we need 3.308 moles of $N_2$. This step is crucial because it utilizes the information provided by the balanced chemical equation. The molar ratio acts as a conversion factor, allowing us to relate the amounts of different substances involved in the reaction. If the ratio were different (say, 2:1), we would need to adjust our calculations accordingly. Always pay close attention to the coefficients in the balanced equation!
Step 3: Convert Moles of $N_2$ to Grams
We know we need 3.308 moles of $N_2$, and we know the molar mass of $N_2$ is 28.02 g/mol. To convert moles to grams, we use the following formula:
Grams = Moles * Molar Mass
So, the grams of $N_2$ needed are:
Grams of $N_2$ = 3.308 moles * 28.02 g/mol ≈ 92.7 grams
The Final Answer: Rounding to the Correct Significant Figures
The question asks us to express our answer to three significant figures. Looking back at our calculations, the least precise number we used was 265 grams (three significant figures). Therefore, our final answer should also have three significant figures. Rounding 92.7 grams to three significant figures gives us 92.7 grams. So, to produce 265 grams of calcium cyanamide, you need approximately 92.7 grams of nitrogen gas.
Conclusion: Stoichiometry in Action
And there you have it! We've successfully calculated the amount of nitrogen needed for this reaction using stoichiometry. We started with a balanced chemical equation, converted grams to moles using molar mass, used the molar ratio from the balanced equation to relate the amounts of reactants and products, and finally converted moles back to grams. Stoichiometry is a powerful tool that allows us to understand and predict the quantitative relationships in chemical reactions. Whether you're a student tackling chemistry problems or a scientist working in a lab, mastering stoichiometry is essential. It's not just about the numbers; it's about understanding the fundamental principles that govern chemical reactions. Keep practicing, and you'll become a stoichiometry whiz in no time!
So, the final answer is 92.7 grams of $N_2$ must be consumed to produce 265 grams of $CaCN_2$.
