Percent Yield Calculation: Mg + I₂ → MgI₂

Hey guys! Today, we're diving into a classic chemistry problem: calculating the percent yield of a reaction. Specifically, we'll look at the reaction between magnesium ($Mg$) and iodine ($I_2$) to form magnesium iodide ($MgI_2$). Let's break it down step-by-step so you can master this concept.

Understanding Percent Yield

Before we jump into the calculations, let's quickly recap what percent yield actually means. In an ideal world, all reactants would convert perfectly into products, giving us a 100% yield. However, in reality, that rarely happens. Percent yield is a measure of the actual amount of product obtained compared to the theoretical amount you should have obtained based on stoichiometry. It's calculated as:

$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Actual Yield: The amount of product you actually get from the reaction (given in the problem).
• Theoretical Yield: The maximum amount of product you could get if everything went perfectly (calculated using stoichiometry).

Problem Statement

Here's the reaction we're working with:

$Mg(s) + I_2(s) \rightarrow MgI_2(s)$

We are given the following information:

• 10.0 g of $Mg$ reacts with 60.0 g of $I_2$.
• 57.84 g of $MgI_2$ is formed (actual yield).

Our goal is to calculate the percent yield of this reaction.

Step-by-Step Solution

Step 1: Calculate the Molar Masses

To work with stoichiometry, we need to convert grams to moles. For this, we need the molar masses of each substance involved. You can find these on the periodic table:

• $Mg$: 24.305 g/mol
• $I_2$: 253.80 g/mol (2 * 126.90 g/mol)
• $MgI_2$: 278.11 g/mol (24.305 + 253.80 g/mol)

Step 2: Determine the Number of Moles of Reactants

Now, let's calculate how many moles of $Mg$ and $I_2$ we have:

• Moles of $Mg = \frac{10.0 \text{ g}}{24.305 \text{ g/mol}} = 0.411 \text{ mol}$
• Moles of $I_2 = \frac{60.0 \text{ g}}{253.80 \text{ g/mol}} = 0.236 \text{ mol}$

Step 3: Identify the Limiting Reactant

The limiting reactant is the reactant that gets used up first, thus limiting the amount of product that can be formed. To find the limiting reactant, we compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. In this case, the balanced equation is already simple: 1 mole of $Mg$ reacts with 1 mole of $I_2$ to produce 1 mole of $MgI_2$.

Divide the number of moles of each reactant by its stoichiometric coefficient (which is 1 in this case):

• For $Mg$: $\frac{0.411 \text{ mol}}{1} = 0.411$
• For $I_2$: $\frac{0.236 \text{ mol}}{1} = 0.236$

Since 0.236 is smaller than 0.411, $I_2$ is the limiting reactant. This means that the amount of $MgI_2$ produced will be limited by the amount of $I_2$ available.

Step 4: Calculate the Theoretical Yield of $MgI_2$

Using the number of moles of the limiting reactant ($I_2$), we can calculate the theoretical yield of $MgI_2$. Since 1 mole of $I_2$ produces 1 mole of $MgI_2$, we will produce 0.236 moles of $MgI_2$.

Now, convert moles of $MgI_2$ to grams:

• Theoretical yield of $MgI_2 = 0.236 \text{ mol} \times 278.11 \text{ g/mol} = 65.64 \text{ g}$

So, theoretically, we should be able to produce 65.64 g of $MgI_2$.

Step 5: Calculate the Percent Yield

Now that we have both the actual yield (57.84 g) and the theoretical yield (65.64 g), we can calculate the percent yield:

$\text{Percent Yield} = \frac{57.84 \text{ g}}{65.64 \text{ g}} \times 100\% = 88.1\%$

Conclusion

The percent yield of the reaction is 88.1%. This means that we obtained 88.1% of the maximum amount of $MgI_2$ that could have been produced based on the amount of limiting reactant ($I_2$) used. Remember, guys, percent yield is a crucial concept in chemistry for evaluating the efficiency of a reaction. Understanding how to calculate it helps us optimize reactions and minimize waste.

Let's recap the key points:

• Molar Masses: Accurate molar masses are essential for converting grams to moles and vice versa.
• Limiting Reactant: Identifying the limiting reactant is crucial because it dictates the maximum amount of product that can be formed.
• Theoretical Yield: The theoretical yield is the maximum amount of product achievable if the reaction goes to completion without any losses.
• Actual Yield: This is the actual amount of product obtained from the experiment.
• Percent Yield Formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

Why is Percent Yield Important?

Understanding percent yield is vital for several reasons, particularly in industrial and research settings. Here’s why:

• Efficiency Assessment: Percent yield indicates how efficiently a chemical reaction converts reactants into products. A high percent yield suggests an efficient reaction with minimal waste, while a low percent yield may indicate significant losses or side reactions.
• Cost Analysis: In industrial processes, knowing the percent yield is crucial for cost analysis. A lower yield means more reactants are needed to produce the desired amount of product, increasing costs. Therefore, optimizing reactions to improve yield can lead to significant cost savings.
• Process Optimization: By understanding the factors that affect percent yield, chemists and engineers can optimize reaction conditions, such as temperature, pressure, and catalyst use, to improve the yield. This is an iterative process that involves experimentation and analysis.
• Troubleshooting: A sudden drop in percent yield can indicate problems in the reaction setup, such as contaminated reactants, equipment malfunctions, or changes in environmental conditions. Monitoring the yield can help identify and troubleshoot these issues quickly.
• Research and Development: In research, percent yield is a key metric for evaluating the success of a new synthetic method. A high yield suggests the method is reliable and efficient, making it more likely to be adopted by others.

Factors Affecting Percent Yield

Several factors can influence the percent yield of a chemical reaction. Being aware of these factors can help in optimizing reactions to achieve the highest possible yield.

• Reaction Conditions: Temperature, pressure, and reaction time play a significant role. Some reactions may require specific temperatures to proceed efficiently, while others may be sensitive to pressure. Insufficient reaction time can also lead to incomplete conversion of reactants.
• Purity of Reactants: Impurities in the reactants can interfere with the reaction, leading to side reactions and a lower yield of the desired product. Using high-purity reactants can help improve the yield.
• Side Reactions: Many chemical reactions can produce unwanted byproducts through side reactions. These side reactions consume reactants and reduce the amount available to form the desired product.
• Equilibrium: Some reactions are reversible and reach an equilibrium state where the forward and reverse reactions occur at the same rate. In such cases, the yield is limited by the position of the equilibrium. Techniques such as removing products or adding excess reactants can shift the equilibrium to favor product formation.
• Losses During Product Isolation: During the process of isolating and purifying the product, some of it may be lost due to handling, filtration, or transfer between containers. Careful technique and optimized purification methods can minimize these losses.
• Stoichiometry: Using the correct stoichiometric ratios of reactants is crucial. If one reactant is present in excess, it may not significantly increase the yield and can complicate the purification process.

Alright, hope this breakdown helps you tackle similar problems with confidence! Keep practicing, and you'll become a pro at calculating percent yields in no time. Happy chemistry, everyone!