MgO Mass For Na2O Production: A Chemistry Guide

Hey chemistry enthusiasts! Let's dive into a classic stoichiometry problem. The question is this: In the reaction $2 NaCl + MgO \rightarrow Na_2O + MgCl_2$, how much mass of Magnesium Oxide (MgO) do we need to get 9.38 grams of Sodium Oxide ($Na_2O$)? This problem involves a chemical reaction, the balanced chemical equation, and the molar masses of the reactants and products. Don't worry; it's not as intimidating as it sounds! We'll break it down into easy-to-follow steps. We will go through all the concepts necessary to solve the problem. Let's get started, shall we?

Understanding the Chemistry Behind the Reaction

First things first, let's get a grip on what's happening chemically. We have Sodium Chloride ($NaCl$) reacting with Magnesium Oxide ($MgO$). This reaction produces Sodium Oxide ($Na_2O$) and Magnesium Chloride ($MgCl_2$). This is a double displacement reaction. It's a straightforward process, but we have to make sure we know the exact quantities involved. The balanced chemical equation is our roadmap, telling us the ratio in which the reactants combine and the products are formed. In our case, the balanced equation tells us that two moles of Sodium Chloride ($NaCl$) react with one mole of Magnesium Oxide ($MgO$) to produce one mole of Sodium Oxide ($Na_2O$) and one mole of Magnesium Chloride ($MgCl_2$). Always start with a balanced equation; otherwise, your calculations will be off, and the whole process will fall apart. Make sure that the number of atoms for each element is equal on both sides of the equation. The coefficients in front of the chemical formulas are crucial. They tell us the mole ratios – the secret ingredient to solving these stoichiometry problems. These coefficients are the bridge that connects the amounts of reactants and products. Without a balanced equation, you're essentially trying to navigate without a map. Understanding the mole ratios is the key to solving all stoichiometry problems. Remember that in chemical reactions, the mass is conserved. So, the total mass of the reactants must equal the total mass of the products. Understanding chemical reactions will help you grasp and solve various chemistry questions.

Step 1: Determine the Molar Mass of Sodium Oxide ($Na_2O$)

Now, let's get to the nitty-gritty of our calculations. The first step is to figure out the molar mass of Sodium Oxide ($Na_2O$). The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). We'll need the periodic table for this. Sodium (Na) has a molar mass of approximately 23.0 g/mol, and Oxygen (O) has a molar mass of about 16.0 g/mol. Since Sodium Oxide ($Na_2O$) has two sodium atoms and one oxygen atom, we calculate its molar mass like this:

• Molar mass of $Na_2O$ = (2 × molar mass of Na) + (1 × molar mass of O)
• Molar mass of $Na_2O$ = (2 × 23.0 g/mol) + (1 × 16.0 g/mol)
• Molar mass of $Na_2O$ = 46.0 g/mol + 16.0 g/mol
• Molar mass of $Na_2O$ = 62.0 g/mol

So, the molar mass of Sodium Oxide ($Na_2O$) is 62.0 g/mol. This means that one mole of Sodium Oxide has a mass of 62.0 grams. This value will be essential in our calculations later on.

Step 2: Convert the Mass of Sodium Oxide ($Na_2O$) to Moles

Next, we'll convert the given mass of Sodium Oxide ($Na_2O$), which is 9.38 grams, into moles. We'll use the molar mass we just calculated (62.0 g/mol) to do this. To convert grams to moles, we use the following formula:

• Moles = Mass / Molar Mass
• Moles of $Na_2O$ = 9.38 g / 62.0 g/mol
• Moles of $Na_2O$ ≈ 0.151 moles

Therefore, 9.38 grams of Sodium Oxide ($Na_2O$) is equivalent to approximately 0.151 moles. We're getting closer to the answer, guys!

Step 3: Use the Mole Ratio from the Balanced Equation

Here's where the balanced equation comes in handy. The equation $2 NaCl + MgO \rightarrow Na_2O + MgCl_2$ tells us that one mole of Magnesium Oxide ($MgO$) is needed to produce one mole of Sodium Oxide ($Na_2O$). This is our mole ratio: 1 mole of $MgO$ : 1 mole of $Na_2O$. Using this ratio, we can determine how many moles of Magnesium Oxide ($MgO$) are needed to produce 0.151 moles of Sodium Oxide ($Na_2O$).

Since the mole ratio is 1:1, the moles of $MgO$ needed are equal to the moles of $Na_2O$ produced.

• Moles of $MgO$ = Moles of $Na_2O$
• Moles of $MgO$ = 0.151 moles

So, we need 0.151 moles of Magnesium Oxide ($MgO$).

Step 4: Calculate the Molar Mass of Magnesium Oxide ($MgO$)

We need the molar mass of Magnesium Oxide ($MgO$) to convert moles of $MgO$ to grams. From the periodic table, Magnesium (Mg) has a molar mass of approximately 24.3 g/mol, and Oxygen (O) has a molar mass of about 16.0 g/mol. Therefore, the molar mass of $MgO$ is:

• Molar mass of $MgO$ = Molar mass of Mg + Molar mass of O
• Molar mass of $MgO$ = 24.3 g/mol + 16.0 g/mol
• Molar mass of $MgO$ = 40.3 g/mol

Step 5: Convert Moles of Magnesium Oxide ($MgO$) to Grams

Finally, we convert the moles of Magnesium Oxide ($MgO$) to grams using its molar mass. We use the formula:

• Mass = Moles × Molar Mass
• Mass of $MgO$ = 0.151 moles × 40.3 g/mol
• Mass of $MgO$ ≈ 6.09 g

Therefore, we need approximately 6.09 grams of Magnesium Oxide ($MgO$) to produce 9.38 grams of Sodium Oxide ($Na_2O$).

Conclusion

There you have it, guys! We've successfully calculated the mass of Magnesium Oxide ($MgO$) needed to produce a certain amount of Sodium Oxide ($Na_2O$). This type of problem is a fundamental skill in chemistry, and by breaking it down step by step, we can solve it with confidence. Always remember to start with a balanced equation and to use molar masses and mole ratios correctly. Keep practicing, and you'll become a stoichiometry pro in no time!