H2S And Hydrogen Volume Calculation: Chemistry Problem

Oct 19, 2025 by SLV Team 55 views

Calculating H₂S and Hydrogen Volume in Chemical Reactions

Hey guys! Let's dive into some interesting chemistry problems today. We're going to tackle calculating the volumes of gases released in chemical reactions, specifically hydrogen sulfide (H₂S) and hydrogen (H₂). This is a classic type of problem in stoichiometry, and understanding the steps involved is crucial for mastering chemical calculations. So, grab your calculators, and let's get started!

Calculating the Volume of H₂S Released

Let's break down the first problem. Our goal is to calculate the volume of H₂S released when iron(II) sulfide (FeS) reacts with hydrochloric acid (HCl). We're given that we have 200 g of FeS reacting with a solution containing 150 g of HCl. Remember, the conditions are standard (STP), which means we can use the molar volume of a gas (22.4 L/mol) once we know the number of moles of H₂S produced. Let's dive into the detailed steps.

First, we need to write the balanced chemical equation for the reaction. This is the foundation for any stoichiometric calculation. The reaction between iron(II) sulfide and hydrochloric acid is a classic acid-base reaction where the sulfide ion (S²⁻) from FeS reacts with the hydrogen ions (H⁺) from HCl to form hydrogen sulfide gas (H₂S). The balanced equation looks like this:

FeS(s) + 2 HCl(aq) → FeCl₂(aq) + H₂S(g)

Now that we have the balanced equation, we can see the stoichiometry of the reaction. One mole of FeS reacts with two moles of HCl to produce one mole of iron(II) chloride (FeCl₂) and one mole of hydrogen sulfide (H₂S). This mole ratio is super important because it tells us how much H₂S we can expect to get from a given amount of FeS and HCl. To figure this out, we need to convert the masses of the reactants (FeS and HCl) into moles. This involves using their molar masses, which we can calculate from the periodic table. The molar mass of FeS is (55.85 + 32.06) g/mol = 87.91 g/mol, and the molar mass of HCl is (1.01 + 35.45) g/mol = 36.46 g/mol.

Using these molar masses, we can convert the given masses of FeS and HCl into moles:

Moles of FeS = (200 g) / (87.91 g/mol) ≈ 2.275 mol
Moles of HCl = (150 g) / (36.46 g/mol) ≈ 4.114 mol

Next, we need to identify the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. From the balanced equation, we know that 1 mole of FeS reacts with 2 moles of HCl. So, we compare the ratio of moles of FeS to moles of HCl:

(Moles of FeS) / (Moles of HCl) = (2.275 mol) / (4.114 mol) ≈ 0.553

The stoichiometric ratio is 1/2 = 0.5. Since the actual ratio (0.553) is greater than the stoichiometric ratio (0.5), it means that we have relatively more FeS than we need to react with all the HCl. Therefore, HCl is the limiting reactant. This means that the amount of H₂S produced will be determined by the amount of HCl available.

Now we can calculate the moles of H₂S produced based on the moles of the limiting reactant (HCl). From the balanced equation, 2 moles of HCl produce 1 mole of H₂S. Therefore, we can set up a proportion:

(Moles of H₂S) / (Moles of HCl) = (1 mol H₂S) / (2 mol HCl)

Moles of H₂S = (1/2) * Moles of HCl = (1/2) * 4.114 mol ≈ 2.057 mol

Finally, we can calculate the volume of H₂S produced at standard conditions (STP). At STP, one mole of any gas occupies 22.4 liters (molar volume). So, we can use this to convert moles of H₂S to volume:

Volume of H₂S = (Moles of H₂S) * (Molar volume at STP)

Volume of H₂S = (2.057 mol) * (22.4 L/mol) ≈ 46.08 L

So, we've calculated that approximately 46.08 liters of H₂S gas are released at standard conditions. This is pretty close to the answer provided (46.03 L), and the slight difference might be due to rounding errors at different steps in the calculation. It's always a good idea to carry as many significant figures as possible throughout your calculations to minimize these errors.

Calculating the Volume of Hydrogen Released

Okay, let's switch gears and move on to the second part of the problem: calculating the volume of hydrogen gas (H₂) released during a reaction. Unfortunately, the original prompt doesn't specify which reaction we're talking about, which is a bit of a bummer. To solve this, we need some more context. Let's assume we're talking about the reaction of a metal with an acid, since that's a common way to produce hydrogen gas. For example, the reaction of zinc (Zn) with hydrochloric acid (HCl) is a classic example. Let's use this as our example and walk through the calculation.

The reaction between zinc and hydrochloric acid produces zinc chloride (ZnCl₂) and hydrogen gas (H₂). The balanced chemical equation for this reaction is:

Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)

Let's assume, for the sake of example, that we have 65.38 g of zinc reacting with excess hydrochloric acid. The molar mass of zinc is 65.38 g/mol. The term "excess" in chemistry problems means that we have more than enough of that reactant to react completely with the other reactant. In this case, we have excess HCl, so zinc will be our limiting reactant. The reason we use this information is because it means the amount of hydrogen gas produced will be determined solely by the amount of zinc we have, simplifying our calculations.

First, we need to convert the mass of zinc into moles:

Moles of Zn = (65.38 g) / (65.38 g/mol) = 1 mol

From the balanced equation, we can see that 1 mole of Zn reacts to produce 1 mole of H₂. So, if we have 1 mole of Zn, we'll produce 1 mole of H₂. This makes the calculation very straightforward because of the 1:1 mole ratio between Zn and H₂ in our balanced chemical equation.

Now that we know the moles of H₂ produced, we can calculate the volume of H₂ at standard conditions (STP). Again, we'll use the molar volume of a gas at STP, which is 22.4 L/mol:

Volume of H₂ = (Moles of H₂) * (Molar volume at STP)

Volume of H₂ = (1 mol) * (22.4 L/mol) = 22.4 L

Therefore, the volume of hydrogen gas released in this example reaction is 22.4 liters at STP. Remember, this is just an example. If we were dealing with a different reaction or different amounts of reactants, the calculation would be similar, but the specific numbers would change. The key is to always start with a balanced chemical equation, identify the limiting reactant, and then use stoichiometry to relate the moles of reactants to the moles of products.

Key Concepts and Takeaways

So, what have we learned today, guys? We've worked through two stoichiometry problems involving gas volumes. Here are some key takeaways to keep in mind:

Balanced Chemical Equations: Always start with a balanced chemical equation. It's the foundation for all stoichiometric calculations. This equation gives you the crucial mole ratios that relate reactants and products.
Molar Mass: You'll need molar masses to convert between mass and moles. Remember to calculate them accurately using the periodic table. The molar mass is essentially the weight of one mole of a substance, expressed in grams per mole (g/mol).
Limiting Reactant: Identify the limiting reactant. It determines the maximum amount of product that can be formed. This reactant gets consumed first, halting the reaction.
Mole Ratios: Use mole ratios from the balanced equation to relate moles of reactants and products. This is the heart of stoichiometry – using the proportions in the balanced equation to predict how much product you'll get.
Molar Volume at STP: At standard conditions (STP), one mole of any gas occupies 22.4 liters. This is your handy conversion factor for gas volume calculations. Remember, STP conditions are defined as 0 degrees Celsius (273.15 K) and 1 atmosphere of pressure.

By mastering these concepts, you'll be well-equipped to tackle a wide range of stoichiometry problems, including those involving gas volumes. Keep practicing, and you'll become a chemistry whiz in no time!

Practice Problems

Want to test your understanding? Try these practice problems:

Calculate the volume of CO₂ produced at STP when 10 grams of methane (CH₄) are completely burned in oxygen.
If 5 grams of magnesium react with excess hydrochloric acid, what volume of hydrogen gas is produced at STP?

Work through these problems step-by-step, using the methods we've discussed, and check your answers. Good luck, and happy calculating!

Remember, stoichiometry can seem tricky at first, but with practice, it becomes much more manageable. So keep at it, guys, and you'll be solving complex chemistry problems in no time! And feel free to ask if anything's unclear – that's what we're here for. Chemistry can be fun, so let's enjoy the process of learning and understanding the world around us! Cheers to mastering the art of chemical calculations!