CuO Reduction: Hydrogen From Magnesium And Hydrochloric Acid

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CuO Reduction: Hydrogen from Magnesium and Hydrochloric Acid

Hey guys! Let's dive into a cool chemistry problem. We're gonna figure out how many grams of copper(II) oxide (CuO) can be reduced by the hydrogen gas produced when 0.2 moles of magnesium react with hydrochloric acid. Sounds fun, right? This is a classic stoichiometry problem, which means we'll be using balanced chemical equations and mole ratios to solve it. Don't worry, it's not as scary as it sounds! We'll break it down step-by-step, and you'll see how everything fits together. Ready to get started? Let's do it!

Step 1: Writing and Balancing the Chemical Equation for Magnesium and Hydrochloric Acid

First things first, we need to know what's happening when magnesium (Mg) and hydrochloric acid (HCl) get together. They react to produce magnesium chloride (MgCl₂) and hydrogen gas (H₂). The chemical equation for this reaction is: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Make sure to balance your equations, guys! Balancing ensures that the number of atoms of each element is the same on both sides of the equation, following the law of conservation of mass. It's like making sure all the ingredients in a recipe are accounted for. So, the balanced equation tells us that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and 1 mole of hydrogen gas. This balanced equation is our roadmap for the rest of the problem. Without it, we wouldn't be able to accurately determine how much hydrogen gas is produced. Understanding the mole ratios is key here! Remember to always double-check your work to avoid any silly mistakes. And, of course, make sure you have your periodic table handy – it's your best friend for these kinds of problems!

We begin by taking the balanced equation and writing down the mole ratio. According to the balanced equation, 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (Hâ‚‚). With this ratio we can now calculate the moles of hydrogen gas produced in this reaction. We start with 0.2 moles of magnesium, and multiply by the mole ratio. The calculation is:

  • Moles of Hâ‚‚ = 0.2 mol Mg × (1 mol Hâ‚‚ / 1 mol Mg) = 0.2 mol Hâ‚‚

So, from the reaction of 0.2 moles of magnesium with hydrochloric acid, we will get 0.2 moles of hydrogen gas.

Step 2: Writing and Balancing the Chemical Equation for the Reduction of Copper(II) Oxide

Next up, we need to understand how hydrogen gas reduces copper(II) oxide (CuO). The reaction here involves hydrogen gas (H₂) reacting with copper(II) oxide (CuO) to produce copper metal (Cu) and water (H₂O). The chemical equation is: CuO(s) + H₂(g) → Cu(s) + H₂O(g). Great, now that's a balanced equation! In this reaction, one mole of CuO reacts with one mole of hydrogen gas to produce one mole of copper and one mole of water. The mole ratio is 1:1, which will make the next step a piece of cake. This balanced equation is essential for calculating the amount of CuO that can be reduced. It tells us the exact relationship between the reactants and products, and without it, we wouldn't know how much copper we can get from the reaction. Pay close attention to the states of matter: solid (s), liquid (l), gas (g), and aqueous (aq). These details are important in some reactions. And always remember to double-check that your equation is balanced because it's a very common mistake! Alright, now let's move on to the next step.

Step 3: Determining the Amount of CuO that Can be Reduced

Now we know that we have 0.2 moles of hydrogen gas. From the balanced equation in step 2 (CuO(s) + H₂(g) → Cu(s) + H₂O(g)), we know that 1 mole of hydrogen gas reduces 1 mole of CuO. Since we have 0.2 moles of H₂, we can reduce 0.2 moles of CuO. Using the mole ratio between hydrogen and copper(II) oxide, we can find out how many moles of CuO will be reduced. We already know from the previous step that we produced 0.2 moles of hydrogen gas. The ratio is:

  • Moles of CuO = 0.2 mol Hâ‚‚ × (1 mol CuO / 1 mol Hâ‚‚) = 0.2 mol CuO

So, 0.2 moles of CuO can be reduced by the hydrogen gas generated in the first reaction. Now let's convert moles to grams, so we can answer the original question.

Step 4: Converting Moles of CuO to Grams

Alright, we've figured out that 0.2 moles of CuO can be reduced. But the question asks for grams, not moles. So, we need to convert moles to grams using the molar mass of CuO. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). To find the molar mass of CuO, we need to add up the atomic masses of copper (Cu) and oxygen (O) from the periodic table. The atomic mass of copper is approximately 63.55 g/mol, and the atomic mass of oxygen is approximately 16.00 g/mol. Therefore, the molar mass of CuO is: 63.55 g/mol (Cu) + 16.00 g/mol (O) = 79.55 g/mol (CuO). Armed with the molar mass of CuO, we can now convert the 0.2 moles of CuO to grams. To do this, we multiply the number of moles by the molar mass: Grams of CuO = 0.2 mol CuO × 79.55 g/mol. And we get: Grams of CuO = 15.91 g. Now we have our answer! This tells us that the hydrogen produced in the reaction can reduce 15.91 grams of copper(II) oxide. It's awesome how we used stoichiometry and balanced equations to solve this problem, right?

Step 5: Final Answer

So, the final answer is that 15.91 grams of CuO can be reduced by the hydrogen gas produced from the reaction of 0.2 moles of magnesium with hydrochloric acid. That wasn't so bad, was it? We've successfully used stoichiometry to solve a real-world chemical problem! You've learned how to: balance chemical equations, use mole ratios, and convert between moles and grams. Keep practicing, and these problems will become second nature to you. Remember, chemistry is all about understanding how substances interact, and stoichiometry gives us the tools to quantify those interactions. Keep up the great work, and happy chemistry-ing, everyone! Don't be afraid to ask for help if you're stuck, and always double-check your work!

Let's recap what we did:

  • We started with the balanced equation for magnesium reacting with hydrochloric acid to get hydrogen gas.
  • We calculated the moles of hydrogen gas produced from the given amount of magnesium.
  • Then, we used the balanced equation for the reduction of copper(II) oxide with hydrogen gas to find the moles of CuO that could be reduced.
  • Finally, we converted the moles of CuO to grams using its molar mass.

And there you have it! A complete solution to the problem. Wasn't that fun? Keep practicing, and you'll become a stoichiometry master in no time! Always remember to balance your equations, understand mole ratios, and convert between units correctly.