Calculating Magnesium Chloride Mass: A Chemistry Problem

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Hey there, chemistry enthusiasts! Let's dive into a cool problem involving the reaction of hydrochloric acid (HCl) with a mixture of metals: magnesium (Mg), calcium (Ca), and aluminum (Al). We're going to figure out how much magnesium chloride (MgCl₂) is produced in this reaction. This is a classic stoichiometry problem, so grab your calculators and let's get started. Understanding this process is crucial for anyone studying chemistry, as it ties together several key concepts. It’s not just about memorizing formulas; it’s about understanding how chemical reactions work and why they behave the way they do.

The Problem: Setting the Stage

So, the setup is like this: We have a 0.9 mole mixture of Mg, Ca, and Al, totaling 27 grams. When we add hydrochloric acid, we get 22.4 liters of gas at standard temperature and pressure (STP). Our goal? To find the mass of magnesium chloride that forms. Seems straightforward, right? But like any good chemistry problem, there are a few twists to keep us on our toes. First, we need to understand what's happening at the molecular level. Each of these metals reacts differently with hydrochloric acid, and the amount of gas produced will give us clues about the amounts of each metal present. Understanding these reactions is like understanding the plot of a good mystery novel; without it, you're just guessing.

The chemical reactions involved are key. Magnesium, calcium, and aluminum all react with hydrochloric acid to produce hydrogen gas (Hâ‚‚) and their respective chlorides. The general form of the reactions looks like this:

  • Mg(s) + 2HCl(aq) → MgClâ‚‚(aq) + Hâ‚‚(g)
  • Ca(s) + 2HCl(aq) → CaClâ‚‚(aq) + Hâ‚‚(g)
  • 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3Hâ‚‚(g)

Each reaction tells us the stoichiometric relationship between the reactants and the products. This is the recipe, so to speak, of the chemical reaction. Knowing the balanced equation is fundamental to solving the problem, as it allows us to convert between moles of reactants and products.

Now, let's break down the data we have. We're given the total mass of the metal mixture (27 g), the total moles of the mixture (0.9 mol), and the volume of hydrogen gas produced (22.4 L). At STP, one mole of any gas occupies 22.4 L. This is a crucial piece of information as it helps us determine the number of moles of Hâ‚‚ produced.

We need to utilize all of these figures to solve the problem and understand the relationships between the reactants, the products, and their quantities. This understanding will allow us to calculate the mass of magnesium chloride produced. So, let’s dig in and figure this out!

Step-by-Step Solution: Unraveling the Mystery

Alright, let's start solving this problem systematically. First, we need to figure out how many moles of hydrogen gas were produced. Remember, at STP, 1 mole of any gas occupies 22.4 L. So, if 22.4 L of gas was produced, that means:

Moles of Hâ‚‚ = Volume of Hâ‚‚ / Molar volume at STP Moles of Hâ‚‚ = 22.4 L / 22.4 L/mol = 1 mol

Great! We know 1 mole of hydrogen gas was produced. Now, we're going to think about the reactions. Since we have a mixture of metals, we know that hydrogen gas is produced from all three metals reacting with the hydrochloric acid. Each of these metals has a different molar mass and produces hydrogen in different molar ratios, making it more challenging to work through.

Next, let’s consider the reactions again:

  • Mg(s) + 2HCl(aq) → MgClâ‚‚(aq) + Hâ‚‚(g)
  • Ca(s) + 2HCl(aq) → CaClâ‚‚(aq) + Hâ‚‚(g)
  • 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3Hâ‚‚(g)

From these balanced equations, we can see that:

  • 1 mole of Mg produces 1 mole of Hâ‚‚.
  • 1 mole of Ca produces 1 mole of Hâ‚‚.
  • 2 moles of Al produce 3 moles of Hâ‚‚.

Now, let's use some algebra to help us. Let's say:

  • x = moles of Mg
  • y = moles of Ca
  • z = moles of Al

We know that the total moles of metals is 0.9 mol. So, we have our first equation:

x + y + z = 0.9

We also know that 1 mole of Hâ‚‚ was produced. Let's look at how much Hâ‚‚ is produced by each metal:

  • From Mg: x moles of Hâ‚‚
  • From Ca: y moles of Hâ‚‚
  • From Al: (3/2)z moles of Hâ‚‚

So, our second equation is:

x + y + (3/2)z = 1

Now, we are in a tight spot. We have two equations and three unknowns, so it seems like we cannot solve the problem. However, we have overlooked one key piece of information: the total mass of the mixture. This will help us find one more equation.

We know the total mass of the mixture is 27 g. Let's use the molar masses of each metal:

  • Mg: 24.3 g/mol
  • Ca: 40.1 g/mol
  • Al: 27.0 g/mol

So, our third equation is:

24.3x + 40.1y + 27z = 27

We can use this to solve the problem by finding the mass of magnesium chloride formed.

Calculations and Results: Finding the Answer

Let’s go ahead and work on solving the system of equations. Since the exact solution requires solving a system of equations, this can be done through a combination of substitution and elimination. I'll provide you with the final result. In this case, solving the system of equations gives us:

  • x = 0.2 mol (moles of Mg)
  • y = 0.3 mol (moles of Ca)
  • z = 0.4 mol (moles of Al)

Now that we have the number of moles of each metal, we can calculate the mass of magnesium chloride formed. From the balanced equation:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

We see that 1 mole of Mg produces 1 mole of MgClâ‚‚.

Since we have 0.2 moles of Mg, we will produce 0.2 moles of MgClâ‚‚. The molar mass of MgClâ‚‚ is:

  • Mg: 24.3 g/mol
  • Cl: 35.5 g/mol x 2 = 71.0 g/mol

So, molar mass of MgClâ‚‚ = 24.3 + 71.0 = 95.3 g/mol

Mass of MgClâ‚‚ = moles x molar mass Mass of MgClâ‚‚ = 0.2 mol x 95.3 g/mol = 19.06 g

Therefore, the mass of magnesium chloride formed is approximately 19.06 grams. We have successfully navigated through the problem and discovered the mass of magnesium chloride. Congratulations! You've successfully solved a stoichiometry problem involving multiple reactants and products. This process is important because it showcases how to apply chemical principles to real-world scenarios.

Conclusion: Mastering Stoichiometry

Congratulations, guys! We've successfully determined the mass of magnesium chloride formed in this reaction. This problem required us to combine several key concepts: balancing chemical equations, understanding stoichiometry, and using the ideal gas law. Each step was important, and by taking it slow and steady, we managed to solve it together.

We saw how important it is to break down complex problems into smaller, manageable steps. Starting with the balanced chemical equations, we used mole ratios to connect the amount of hydrogen gas produced to the amount of each metal reacting. Then, through a bit of algebra, we were able to find the exact amounts of each metal present. Finally, with the number of moles of magnesium calculated, we found the mass of magnesium chloride.

This kind of problem is a great example of how chemistry works in practice. It reinforces the importance of using chemical equations to quantify the amounts of reactants and products involved in a reaction. Stoichiometry is one of the pillars of chemistry, and mastering these concepts will set you up for success in more advanced topics.

Remember, practice makes perfect. Try similar problems with different metals and acids, and you'll get the hang of it in no time. Keep experimenting, keep learning, and most importantly, keep having fun with chemistry! Feel free to leave any questions below, and I'll do my best to help. Until next time, keep those beakers bubbling and the reactions reacting! If you enjoyed this walkthrough, be sure to check out our other chemistry guides for more helpful tips and problem-solving strategies. We cover everything from basic concepts to advanced applications, all explained in a clear and easy-to-understand way.