Calculate Hydrogen For Ammonia Production

Hey chemistry dudes and dudettes! Today, we're diving into a classic stoichiometry problem that's super important in understanding how chemists make things. We're talking about the Haber-Bosch process, where nitrogen and hydrogen team up to create ammonia. It's a pretty big deal, guys, because ammonia is a key ingredient in fertilizers, which, you know, helps us grow all the food we eat. So, let's get our nerd on and figure out the exact amount of hydrogen we need to make a specific amount of ammonia. This isn't just about passing a test; it's about understanding the quantitative relationships in chemical reactions, which is like the secret sauce of chemistry. We'll break down the balanced chemical equation and use molar masses to convert moles to grams. Get ready to flex those chemistry muscles!

The Heart of the Reaction: Nitrogen Meets Hydrogen

The balanced chemical equation for the production of ammonia is absolutely crucial here, guys. It tells us the precise ratio in which our reactants combine and products form. So, let's look at it: $N_2(g) + 3 H_2(g) ightarrow 2 NH_3(g)$. What this equation is screaming at us is that one mole of nitrogen gas ($N_2$) reacts with three moles of hydrogen gas ($H_2$) to produce two moles of ammonia ($NH_3$). This 1:3:2 ratio is our golden ticket to solving stoichiometry problems. Without a balanced equation, we'd be flying blind! It's like trying to bake a cake without a recipe – you might end up with something edible, but probably not what you intended. This equation isn't just a bunch of letters and numbers; it's a recipe for chemical change, showing us exactly how much of each ingredient is needed and how much product we can expect. So, remember this ratio: for every 2 moles of ammonia we want to make, we'll need 3 moles of hydrogen. This relationship is the foundation for all our calculations.

From Moles to Mass: The Gram Calculation

Now, the question asks us how many grams of hydrogen are needed to produce 15.0 mol of ammonia. Notice that the question gives us the amount of product in moles, but it asks for the amount of reactant in grams. This means we need to use our trusty molar masses to convert between moles and grams. First things first, we need the molar mass of hydrogen ($H_2$). Hydrogen is element number 1 on the periodic table, and its atomic mass is approximately 1.008 g/mol. Since hydrogen exists as a diatomic molecule ($H_2$), its molar mass is about $2 imes 1.008$ g/mol, which is roughly 2.016 g/mol. We'll use this value to convert the moles of hydrogen we calculate into grams. This conversion step is vital because, in a real lab, we measure out chemicals by mass (grams) or volume, not usually by moles directly. So, understanding how to move between these units is a fundamental skill for any aspiring chemist. It’s all about making the chemistry practical and applicable to real-world scenarios.

Step-by-Step: Unraveling the Calculation

Alright, let's break down this calculation step-by-step, making sure we don't miss a beat. We know we want to produce 15.0 mol of ammonia ($NH_3$). Our balanced equation tells us that 2 moles of $NH_3$ are produced from 3 moles of $H_2$. This gives us our first conversion factor. So, to find out how many moles of $H_2$ we need for 15.0 mol of $NH_3$, we can set up a ratio:

(15.0 mol $NH_3$) $ imes$ (3 mol $H_2$ / 2 mol $NH_3$)

Notice how the 'mol $NH_3$' units cancel out, leaving us with 'mol $H_2$', which is exactly what we want. This calculation gives us: 22.5 mol $H_2$. So, we need 22.5 moles of hydrogen gas to produce 15.0 moles of ammonia. Now, the question wants the answer in grams. This is where our molar mass of $H_2$ comes in. We calculated it to be approximately 2.016 g/mol. So, to convert moles of $H_2$ to grams of $H_2$, we multiply:

(22.5 mol $H_2$) $ imes$ (2.016 g $H_2$ / 1 mol $H_2$)

Again, the 'mol $H_2$' units cancel, leaving us with grams of $H_2$. Performing this multiplication gives us approximately 45.36 grams of $H_2$. So, to produce 15.0 mol of ammonia, you'll need about 45.36 grams of hydrogen. Pretty neat, huh? This process highlights the power of stoichiometry in predicting the exact quantities needed for chemical reactions, ensuring efficiency and minimizing waste in industrial processes.

Conclusion: The Answer Revealed!

So, after all that calculation and breakdown, we've arrived at our answer. We needed to find out how many grams of hydrogen ($H_2$) are required to produce 15.0 mol of ammonia ($NH_3$) based on the reaction $N_2(g) + 3 H_2(g) ightarrow 2 NH_3(g)$. Using the stoichiometric ratio from the balanced equation, we determined that we need 22.5 mol of $H_2$ for 15.0 mol of $NH_3$. Then, by using the molar mass of $H_2$ (approximately 2.016 g/mol), we converted those moles into grams. The final calculation yielded 45.36 grams of $H_2$. Looking at the options provided (A. 15.0 g, B. 22.5 g, C. 45.0 g, D. 90.0 g), our calculated value is closest to 45.0 g. Therefore, the correct answer is C. 45.0 g. This problem is a fantastic example of how stoichiometry works, allowing us to quantitatively link reactants and products in a chemical reaction. It's this kind of precision that makes chemistry so powerful and applicable in so many different fields, from pharmaceuticals to material science. Keep practicing these calculations, guys, and you'll be a stoichiometry whiz in no time!