Magnesium Hydroxide Production Calculation From Magnesium Chloride And Sodium Hydroxide Reaction

Hey guys! Let's dive into a cool chemistry problem together. We're going to figure out how much magnesium hydroxide (${Mg(OH)_2}$) is produced when we react magnesium chloride (${MgCl_2}$) with excess sodium hydroxide (${NaOH}$). This is a classic stoichiometry problem, and we'll break it down step-by-step so it's super easy to follow. So, grab your calculators and let's get started!

The Chemical Reaction: A Quick Recap

Before we jump into the calculations, let's quickly review the chemical reaction we're dealing with. The reaction between sodium hydroxide and magnesium chloride produces sodium chloride (${NaCl}$) and magnesium hydroxide, as shown in the balanced equation:

${2 NaOH(aq) + MgCl_2(aq) \rightarrow 2 NaCl(aq) + Mg(OH)_2(s)}$

This balanced equation is our roadmap. It tells us the exact ratio in which the reactants combine and the products are formed. In this case, it tells us that 2 moles of sodium hydroxide react with 1 mole of magnesium chloride to produce 2 moles of sodium chloride and 1 mole of magnesium hydroxide. This mole ratio is crucial for our calculations. Think of it like a recipe – if you don't have the right proportions, your final dish (or in this case, product) won't be right!

Stoichiometry: The Heart of the Calculation

Stoichiometry is a fancy word, but all it really means is using the relationships between reactants and products in a chemical reaction to calculate amounts. We're going to use the stoichiometric coefficients from our balanced equation to convert between moles of different substances. Remember, the coefficients in front of each chemical formula in the balanced equation represent the number of moles involved in the reaction. For example, the "2" in front of "NaOH" means that 2 moles of sodium hydroxide are involved for every 1 mole of magnesium chloride. This mole ratio acts as a conversion factor, allowing us to move from the quantity of one substance to another. It’s like saying, “For every two slices of bread, I need one slice of cheese to make a sandwich.” The ratio is what connects the ingredients!

In this problem, we are starting with a known mass of magnesium chloride (4.6 grams) and we want to find out the mass of magnesium hydroxide produced. To do this, we’ll need to go through a few steps:

1. Convert the mass of magnesium chloride to moles.
2. Use the mole ratio from the balanced equation to find the moles of magnesium hydroxide produced.
3. Convert the moles of magnesium hydroxide to grams.

Let's tackle each of these steps one by one.

Step 1: Converting Grams of Magnesium Chloride to Moles

The first step in our journey is converting the given mass of magnesium chloride (4.6 grams) into moles. Why moles? Because the balanced equation speaks in terms of moles, not grams. Moles are the chemist’s counting unit – it’s a way of standardizing the number of particles we’re dealing with, even though those particles (atoms, molecules, ions) have different masses.

To convert grams to moles, we need the molar mass of magnesium chloride (${MgCl_2}$). The molar mass is the mass of one mole of a substance, and it's calculated by adding up the atomic masses of all the atoms in the chemical formula. You can find the atomic masses on the periodic table. Here's how we calculate the molar mass of ${MgCl_2}$:

• Magnesium (Mg): 24.31 g/mol
• Chlorine (Cl): 35.45 g/mol (but we have two chlorine atoms, so we multiply by 2)

Molar mass of ${MgCl_2}$ = 24.31 g/mol + (2 × 35.45 g/mol) = 95.21 g/mol

Now that we have the molar mass, we can use it as a conversion factor to convert grams of ${MgCl_2}$ to moles. We'll use the following formula: ${Moles = \frac{Mass}{Molar Mass}}$

Plugging in our values: ${Moles of MgCl_2 = \frac{4.6 \text{ grams}}{95.21 \text{ g/mol}} = 0.0483 \text{ moles}}$

So, we have 0.0483 moles of magnesium chloride. We've successfully converted grams to moles – high five! Now we're one step closer to our final answer.

Step 2: Using the Mole Ratio to Find Moles of Magnesium Hydroxide

This is where the balanced equation really shines! Remember, the coefficients in the balanced equation give us the mole ratio between the reactants and products. We want to find the moles of magnesium hydroxide (${Mg(OH)_2}$) produced, and we know the moles of magnesium chloride (${MgCl_2}$) we started with.

Looking back at the balanced equation:

${2 NaOH(aq) + MgCl_2(aq) \rightarrow 2 NaCl(aq) + Mg(OH)_2(s)}$

We see that 1 mole of ${MgCl_2}$ produces 1 mole of ${Mg(OH)_2}$. This is a simple 1:1 mole ratio. This ratio is our key to unlocking the moles of magnesium hydroxide. For every one mole of magnesium chloride that reacts, one mole of magnesium hydroxide is formed. So, if we know the moles of magnesium chloride, we instantly know the moles of magnesium hydroxide produced. In this case, since the ratio is 1:1, the moles of ${Mg(OH)_2}$ produced will be the same as the moles of ${MgCl_2}$ reacted.

Using the mole ratio as a conversion factor, we can set up the following calculation: ${Moles of Mg(OH)_2 = Moles of MgCl_2 × \frac{1 \text{ mole } Mg(OH)_2}{1 \text{ mole } MgCl_2}}$

Plugging in the value we calculated in Step 1: ${Moles of Mg(OH)_2 = 0.0483 \text{ moles } MgCl_2 × \frac{1 \text{ mole } Mg(OH)_2}{1 \text{ mole } MgCl_2} = 0.0483 \text{ moles}}$

Therefore, 0.0483 moles of magnesium hydroxide are produced. See? The mole ratio made this step super straightforward! We've now found the moles of magnesium hydroxide – we’re on the home stretch!

Step 3: Converting Moles of Magnesium Hydroxide to Grams

We've calculated the moles of magnesium hydroxide produced, but the question likely asks for the mass in grams. So, our final step is to convert moles back to grams. Guess what we'll use? That's right, the molar mass! This is the reverse process of what we did in Step 1.

First, we need to calculate the molar mass of magnesium hydroxide (${Mg(OH)_2}$). Again, we'll use the periodic table to find the atomic masses:

• Magnesium (Mg): 24.31 g/mol
• Oxygen (O): 16.00 g/mol (but we have two oxygen atoms, so we multiply by 2)
• Hydrogen (H): 1.01 g/mol (but we have two hydrogen atoms, so we multiply by 2)

Molar mass of ${Mg(OH)_2}$ = 24.31 g/mol + (2 × 16.00 g/mol) + (2 × 1.01 g/mol) = 58.33 g/mol

Now that we have the molar mass, we can use it to convert moles of ${Mg(OH)_2}$ to grams. We'll use a slightly rearranged version of the formula we used in Step 1: ${Mass = Moles × Molar Mass}$

Plugging in our values: ${Mass of Mg(OH)_2 = 0.0483 \text{ moles} × 58.33 \text{ g/mol} = 2.82 \text{ grams}}$

Drumroll, please! We've done it! We've calculated that 2.82 grams of magnesium hydroxide are produced in this reaction.

The Final Answer: 2.82 Grams of Magnesium Hydroxide

So, there you have it! When excess sodium hydroxide is added to a solution containing 4.6 grams of magnesium chloride, 2.82 grams of magnesium hydroxide are produced. We tackled this stoichiometry problem step-by-step, converting grams to moles, using the mole ratio from the balanced equation, and then converting back to grams. Each step built upon the previous one, and by carefully following the process, we arrived at the correct answer.

Key Takeaways and Stoichiometry Superpowers

• Balanced Chemical Equations are Key: The balanced equation provides the crucial mole ratios needed for stoichiometric calculations. It's the foundation of everything we do.
• Moles are the Chemist's Unit: Converting masses to moles is essential for working with chemical reactions because the balanced equation expresses the reaction in terms of moles.
• Molar Mass is Your Friend: The molar mass is the bridge between grams and moles, allowing you to convert between these units.
• Step-by-Step Approach: Breaking down complex problems into smaller, manageable steps makes them much easier to solve.

By mastering these concepts, you'll unlock your stoichiometry superpowers and be able to solve a wide range of chemical calculation problems! Keep practicing, and you'll become a stoichiometry whiz in no time. Keep up the awesome work, chemistry rockstars!