Limiting Reagent And Percent Yield Practice

The heart of chemistry lies in understanding how much of a substance you can create from a reaction, and that's where limiting reagents and percent yield come into play. These concepts are crucial for accurately predicting and optimizing chemical reactions in the lab, as well as in industrial settings. Mastering them allows you to calculate the maximum amount of product possible and evaluate the efficiency of the chemical reactions you conduct.

What is a Limiting Reagent?

In every chemical reaction, reactants combine in specific proportions as dictated by the balanced chemical equation. However, we often don't use the exact, perfect amounts of each reactant. When this happens, one reactant will be completely consumed before the others. This "completely used up" reactant is known as the limiting reagent.

Think of it like making sandwiches. Let's say you want to make turkey sandwiches, and you have:

• 10 slices of bread
• 7 slices of turkey
• 1 bottle of mustard

Even though you have plenty of bread and mustard, you can only make 3 sandwiches with turkey because you only have 7 slices. The turkey is the limiting reagent in this sandwich-making "reaction." You'll run out of turkey before you run out of bread or mustard.

The limiting reagent limits the amount of product (turkey sandwiches) that can be formed. The other reactants (bread and mustard, in our analogy) are in excess.

Identifying the Limiting Reagent

To identify the limiting reagent, you need to:

1. Write a balanced chemical equation: This equation provides the stoichiometric ratios between reactants and products.
2. Convert the mass of each reactant to moles: Moles are the standard unit of measurement for the amount of substance in chemistry. Use the molar mass of each reactant to perform this conversion. Molar mass is found on the periodic table.
3. Divide the number of moles of each reactant by its stoichiometric coefficient from the balanced equation: This step normalizes the number of moles based on the reaction's stoichiometry.
4. Compare the results: The reactant with the smallest value after the division is the limiting reagent.

Let's illustrate this with a chemical example:

Example: Suppose you react 10.0 g of nitrogen gas (N2) with 3.0 g of hydrogen gas (H2) to produce ammonia (NH3). Which is the limiting reagent?

1. Balanced equation: N2(g) + 3H2(g) → 2NH3(g)
2. Convert to moles:
   • Moles of N2 = 10.0 g / 28.02 g/mol = 0.357 mol
   • Moles of H2 = 3.0 g / 2.02 g/mol = 1.49 mol
3. Divide by stoichiometric coefficient:
   • N2: 0.357 mol / 1 = 0.357
   • H2: 1.49 mol / 3 = 0.497
4. Compare: 0.357 < 0.497, therefore, nitrogen (N2) is the limiting reagent.

What is Percent Yield?

In an ideal world, every reaction would proceed perfectly, converting all of the limiting reagent into the desired product. However, this rarely happens. There are many reasons why a reaction might not yield the maximum possible amount of product, including:

• Side reactions: Reactants might participate in unintended reactions, forming byproducts instead of the desired product.
• Incomplete reactions: The reaction may not go to completion, leaving some of the limiting reagent unreacted.
• Losses during purification: When separating the product from the reaction mixture, some product can be lost during filtration, distillation, or other purification steps.

To quantify the efficiency of a reaction, we use percent yield. Percent yield compares the actual amount of product obtained (the actual yield) to the maximum amount of product that could have been obtained based on the amount of limiting reagent used (the theoretical yield).

The formula for percent yield is:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

• Actual Yield: The mass of the pure, isolated product you actually obtain from the reaction in the lab.
• Theoretical Yield: The maximum mass of product that can be produced based on the stoichiometry of the reaction and the amount of the limiting reagent used. This is calculated.

Calculating Theoretical Yield

To calculate the theoretical yield:

1. Identify the limiting reagent (as described above).
2. Determine the mole ratio between the limiting reagent and the desired product from the balanced chemical equation.
3. Calculate the moles of product that can be formed from the moles of limiting reagent using the mole ratio.
4. Convert moles of product to grams of product using the molar mass of the product. This is your theoretical yield.

Example (Continuing from the previous example): If the reaction of 10.0 g of N2 with 3.0 g of H2 produced 8.0 g of NH3, what is the percent yield? We already determined that N2 is the limiting reagent.

1. Mole ratio: From the balanced equation (N2(g) + 3H2(g) → 2NH3(g)), 1 mole of N2 produces 2 moles of NH3.
2. Moles of product: Since we started with 0.357 moles of N2, the theoretical number of moles of NH3 produced is 0.357 mol N2 * (2 mol NH3 / 1 mol N2) = 0.714 mol NH3.
3. Convert to grams: The molar mass of NH3 is 17.03 g/mol. Therefore, the theoretical yield is 0.714 mol NH3 * 17.03 g/mol = 12.16 g NH3.
4. Calculate Percent Yield: Actual yield = 8.0 g. Percent yield = (8.0 g / 12.16 g) * 100% = 65.8%.

Practice Problems: Limiting Reagent and Percent Yield

Now, let's solidify your understanding with some practice problems.

Problem 1:

When 5.00 g of copper (II) chloride (CuCl2) react with 4.00 g of sodium nitrate (NaNO3), copper (II) nitrate (Cu(NO3)2) and sodium chloride (NaCl) are formed.

CuCl2(aq) + 2 NaNO3(aq) → Cu(NO3)2(aq) + 2 NaCl(aq)

(a) Which is the limiting reagent? (b) How many grams of copper (II) nitrate (Cu(NO3)2) are formed? (c) If 6.80 g of copper (II) nitrate (Cu(NO3)2) are actually obtained, what is the percent yield?

Solution:

(a) Limiting Reagent:

1. Convert to moles:
   • Moles of CuCl2 = 5.00 g / 134.45 g/mol = 0.0372 mol
   • Moles of NaNO3 = 4.00 g / 85.00 g/mol = 0.0471 mol
2. Divide by stoichiometric coefficient:
   • CuCl2: 0.0372 mol / 1 = 0.0372
   • NaNO3: 0.0471 mol / 2 = 0.0236
3. Compare: 0.0372 > 0.0236, therefore, sodium nitrate (NaNO3) is the limiting reagent.

(b) Theoretical Yield of Cu(NO3)2:

1. Mole ratio: From the balanced equation, 2 moles of NaNO3 produce 1 mole of Cu(NO3)2.
2. Moles of product: Since NaNO3 is the limiting reagent, moles of Cu(NO3)2 produced = 0.0471 mol NaNO3 * (1 mol Cu(NO3)2 / 2 mol NaNO3) = 0.0236 mol Cu(NO3)2
3. Convert to grams: The molar mass of Cu(NO3)2 is 187.56 g/mol. Therefore, the theoretical yield is 0.0236 mol Cu(NO3)2 * 187.56 g/mol = 4.43 g Cu(NO3)2

(c) Percent Yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Percent Yield = (6.80 g / 4.43 g) x 100% = 153.5%

Wait, what?! How can the percent yield be over 100%? This result indicates that the product was not completely pure. The actual yield of 6.80 g likely includes impurities, such as unreacted starting materials or byproducts, that increased the mass of the isolated solid. Percent yields exceeding 100% are a sign of experimental error and indicate the need to carefully purify the product.

Problem 2:

Consider the reaction of iron (III) oxide (Fe2O3) with carbon monoxide (CO) to produce iron (Fe) and carbon dioxide (CO2).

Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

If you start with 25.0 g of Fe2O3 and 30.0 g of CO:

(a) Which is the limiting reagent? (b) What is the theoretical yield of iron (Fe) in grams? (c) If the actual yield of iron is 15.0 g, what is the percent yield?

Solution:

(a) Limiting Reagent:

1. Convert to moles:
   • Moles of Fe2O3 = 25.0 g / 159.69 g/mol = 0.157 mol
   • Moles of CO = 30.0 g / 28.01 g/mol = 1.07 mol
2. Divide by stoichiometric coefficient:
   • Fe2O3: 0.157 mol / 1 = 0.157
   • CO: 1.07 mol / 3 = 0.357
3. Compare: 0.157 < 0.357, therefore, iron (III) oxide (Fe2O3) is the limiting reagent.

(b) Theoretical Yield of Iron (Fe):

1. Mole ratio: From the balanced equation, 1 mole of Fe2O3 produces 2 moles of Fe.
2. Moles of product: Since Fe2O3 is the limiting reagent, moles of Fe produced = 0.157 mol Fe2O3 * (2 mol Fe / 1 mol Fe2O3) = 0.314 mol Fe
3. Convert to grams: The molar mass of Fe is 55.845 g/mol. Therefore, the theoretical yield is 0.314 mol Fe * 55.845 g/mol = 17.54 g Fe

(c) Percent Yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Percent Yield = (15.0 g / 17.54 g) x 100% = 85.5%

Problem 3:

The reaction between potassium iodide (KI) and lead (II) nitrate (Pb(NO3)2) produces lead (II) iodide (PbI2) and potassium nitrate (KNO3).

2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)

If you start with 15.0 g of KI and 20.0 g of Pb(NO3)2:

(a) Which is the limiting reagent? (b) What is the theoretical yield of lead (II) iodide (PbI2) in grams? (c) If you isolate 18.0 g of PbI2, what is the percent yield?

Solution:

(a) Limiting Reagent:

1. Convert to moles:
   • Moles of KI = 15.0 g / 166.00 g/mol = 0.0904 mol
   • Moles of Pb(NO3)2 = 20.0 g / 331.2 g/mol = 0.0604 mol
2. Divide by stoichiometric coefficient:
   • KI: 0.0904 mol / 2 = 0.0452
   • Pb(NO3)2: 0.0604 mol / 1 = 0.0604
3. Compare: 0.0452 < 0.0604, therefore, potassium iodide (KI) is the limiting reagent.

(b) Theoretical Yield of Lead (II) Iodide (PbI2):

1. Mole ratio: From the balanced equation, 2 moles of KI produce 1 mole of PbI2.
2. Moles of product: Since KI is the limiting reagent, moles of PbI2 produced = 0.0904 mol KI * (1 mol PbI2 / 2 mol KI) = 0.0452 mol PbI2
3. Convert to grams: The molar mass of PbI2 is 461.01 g/mol. Therefore, the theoretical yield is 0.0452 mol PbI2 * 461.01 g/mol = 20.84 g PbI2

(c) Percent Yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Percent Yield = (18.0 g / 20.84 g) x 100% = 86.4%

Tips for Success

• Double-check your calculations: Errors in molar mass calculations or stoichiometric ratios will lead to incorrect results.
• Pay attention to units: Always include units in your calculations and make sure they cancel out correctly.
• Use significant figures: Report your final answer with the appropriate number of significant figures.
• Understand the limitations: Percent yield provides a measure of reaction efficiency, but it doesn't tell the whole story. It's important to also consider factors such as the cost of reactants, the toxicity of byproducts, and the environmental impact of the reaction when evaluating a chemical process.

Conclusion

Limiting reagent and percent yield are fundamental concepts in chemistry that allow us to predict and assess the outcome of chemical reactions. By mastering these calculations, you can optimize reactions to maximize product yield and minimize waste. These skills are not only essential for success in the chemistry lab, but also for understanding and improving chemical processes in a wide range of industries. So, keep practicing, and you'll become a true chemical master!