Solving Lithium And Magnesium Oxide Mixture Problem Step-by-Step

by Axel Sørensen 65 views

Hey guys! Have you ever stumbled upon a chemistry problem that seemed like a real head-scratcher? Well, today, we're diving deep into one of those intriguing scenarios – a problem involving the reaction of lithium and magnesium with oxygen. Specifically, we're tackling a situation where the mass of the solid product doubles after the reaction. Sounds fascinating, right? Let's break it down together, step by step, and make sure we not only solve it but also truly understand the chemistry behind it.

Delving into the Problem: Lithium and Magnesium Oxide Formation

So, here's the gist of the problem: We have a mixture of lithium and magnesium reacting with oxygen. This reaction leads to the formation of oxides, and here's the kicker – the mass of the solid stuff we end up with is double what we started with. We know we had 1.2 grams of magnesium in the mix, and our mission, should we choose to accept it (and we do!), is to figure out how much lithium, in moles, was in the original mixture. The possible answers are laid out before us: A) 0.8, B) 0.1, C) 0.4, and D) 0.6. Time to put on our thinking caps and get to work!

Understanding the Chemical Reactions at Play

First things first, let's jot down the balanced chemical equations for the reactions happening here. This is crucial because it gives us the blueprint of how lithium and magnesium interact with oxygen. Remember, chemistry is all about the ratios, so getting these equations right is our foundation.

  • For lithium (Li) reacting with oxygen (O₂), we get lithium oxide (Li₂O). The balanced equation looks like this:

    4Li + O₂ → 2Li₂O

    This tells us that 4 moles of lithium react with 1 mole of oxygen to produce 2 moles of lithium oxide. Keep this ratio in mind; it's gold!

  • Next up, magnesium (Mg) reacts with oxygen (O₂) to form magnesium oxide (MgO). The balanced equation is:

    2Mg + O₂ → 2MgO

    Here, 2 moles of magnesium react with 1 mole of oxygen to give us 2 moles of magnesium oxide.

With these equations in hand, we're armed and ready to tackle the problem head-on. These equations are the key to unlocking the relationship between the reactants and the products, and they'll guide us in figuring out the amount of lithium involved.

Calculating Moles of Magnesium: Our Starting Point

We're given that the mass of magnesium (Mg) is 1.2 grams. To use this information in our balanced equations, we need to convert grams to moles. Remember the formula: moles = mass / molar mass. The molar mass of magnesium is approximately 24 g/mol (you can find this on the periodic table, guys!). So,

Moles of Mg = 1.2 g / 24 g/mol = 0.05 moles

This means we had 0.05 moles of magnesium in our initial mixture. This is a concrete value we can use to start piecing together the rest of the puzzle.

Decoding the Mass Increase: Oxygen's Role

The problem states that the mass of the solid doubled after the reaction. This increase in mass is entirely due to the oxygen that has combined with the lithium and magnesium to form the oxides. Let's think about this logically. If the mass doubled, that means the mass of oxygen that reacted is equal to the initial mass of the lithium and magnesium mixture. This is a critical understanding.

Let's denote the initial mass of the lithium as 'x' grams. Then, the total initial mass of the mixture is (x + 1.2) grams (remember, we had 1.2 grams of magnesium). Since the mass doubled, the mass of oxygen that reacted is also (x + 1.2) grams. Now, let's convert this mass of oxygen to moles. The molar mass of O₂ is approximately 32 g/mol. So,

Moles of O₂ = (x + 1.2) g / 32 g/mol

This expression gives us the total moles of oxygen that reacted with both lithium and magnesium. It's like we're building a bridge between the mass increase and the chemical quantities involved. Keep this equation in your toolkit; we'll use it shortly.

Connecting the Dots: Moles of Oxygen in Each Reaction

Now, let's figure out how much oxygen reacted with each element separately. We already know we have 0.05 moles of magnesium. Looking back at the balanced equation for magnesium (2Mg + O₂ → 2MgO), we see that 2 moles of Mg react with 1 mole of O₂. Therefore, the moles of O₂ that reacted with magnesium are:

Moles of O₂ (with Mg) = 0.05 moles Mg * (1 mole O₂ / 2 moles Mg) = 0.025 moles

So, 0.025 moles of oxygen were used up in the reaction with magnesium. Now, we can figure out how much oxygen reacted with lithium. If we subtract the moles of O₂ used by magnesium from the total moles of O₂ that reacted, we'll have our answer.

Let's say 'y' is the moles of lithium. The balanced equation for lithium (4Li + O₂ → 2Li₂O) tells us that 4 moles of Li react with 1 mole of O₂. So, the moles of O₂ that reacted with lithium are:

Moles of O₂ (with Li) = y moles Li * (1 mole O₂ / 4 moles Li) = y / 4 moles

Now we can express the total moles of O₂ that reacted as the sum of the moles used by magnesium and lithium:

Total moles of O₂ = Moles of O₂ (with Mg) + Moles of O₂ (with Li)
(x + 1.2) / 32 = 0.025 + y / 4

This equation is a big step forward. It links the mass of lithium ('x') to the moles of lithium ('y') and the moles of oxygen involved. It's like we're weaving a tapestry of chemical relationships!

The Final Piece: Relating Mass and Moles of Lithium

We're almost there, guys! We have an equation with 'x' (mass of lithium) and 'y' (moles of lithium). To solve this, we need one more connection: the relationship between mass and moles. We know that moles = mass / molar mass. The molar mass of lithium is approximately 7 g/mol. So,

y = x / 7

Now we have a system of two equations:

  1. (x + 1.2) / 32 = 0.025 + y / 4
  2. y = x / 7

We can substitute the second equation into the first to solve for 'x':

(x + 1.2) / 32 = 0.025 + (x / 7) / 4
(x + 1.2) / 32 = 0.025 + x / 28

Now, let's solve for 'x'. Multiply both sides by the least common multiple of 32 and 28, which is 224:

7(x + 1.2) = 224 * 0.025 + 8x
7x + 8.4 = 5.6 + 8x

Subtract 7x from both sides and subtract 5.6 from both sides:

8.  4 - 5.6 = 8x - 7x
    2.  8 = x

So, x = 2.8 grams. Now we can find the moles of lithium (y) using the equation y = x / 7:

y = 2.8 g / 7 g/mol = 0.4 moles

Conclusion: Cracking the Code

And there we have it! The amount of lithium in the original mixture is 0.4 moles. So, the correct answer is C) 0.4. Woohoo! We did it, guys!

This problem was a fantastic journey through stoichiometry, balanced equations, and molar mass calculations. The key takeaway here is the importance of breaking down complex problems into smaller, manageable steps. By carefully considering each piece of information and how it relates to the chemical reactions, we were able to navigate through the problem and arrive at the correct solution.

Remember, chemistry might seem daunting at first, but with a systematic approach and a solid understanding of the fundamentals, you can conquer any challenge. Keep practicing, keep exploring, and most importantly, keep having fun with chemistry!