Problem 214 - Introductory Chemistry Atoms First

Question

Suppose \(200.0 \mathrm{~mL}\) of a \(2.50 \mathrm{M}\) solution of sodium hydroxide is combined with \(100.0 \mathrm{~mL}\) of a \(1.50 \mathrm{M}\) solution of iron(III) nitrate. (a) Write a net ionic equation for the formation of the expected precipitate. (b) What is the theoretical yield of precipitate in grams? (c) Suppose only \(10.95 \mathrm{~g}\) of precipitate is isolated. What is the percent yield for the reaction? (d) What is the molar concentration of the excess reactant in the combined solution?

Step-by-Step Solution

Answer

(a) The net ionic equation for the formation of the expected precipitate is: \(2 \mathrm{OH^-} + \mathrm{Fe^{3+}} \rightarrow \mathrm{Fe(OH)_3}\) (b) The theoretical yield of precipitate is 15.9 g. (c) The percent yield for the reaction is ≈ 68.87%. (d) The molar concentration of the excess reactant (NaOH) in the combined solution is ≈ 0.667 M.

Step 1: Write the balanced chemical equation and net ionic equation

First, we need to write the balanced chemical equation for the reaction: \(2 \mathrm{NaOH} + \mathrm{Fe(NO_3)_3} \rightarrow 3 \mathrm{NaNO_3} + \mathrm{Fe(OH)_3}\) Now, we should write the net ionic equation by removing spectator ions (the ions that don't participate in the reaction): \(2 \mathrm{OH^-} + \mathrm{Fe^{3+}} \rightarrow \mathrm{Fe(OH)_3}\)

Step 2: Calculate the theoretical yield of the precipitate

To find the theoretical yield, we start by determining the limiting reactant. We have 200.0 mL of 2.50 M NaOH solution and 100.0 mL of 1.50 M Fe(NO3)3 solution. Moles of NaOH = Molarity × Volume = 2.50 M × 0.200 L = 0.500 mol Moles of Fe(NO3)3 = Molarity × Volume = 1.50 M × 0.100 L = 0.150 mol From the balanced chemical equation, we see that 2 moles of NaOH react with 1 mole of Fe(NO3)3. The ratio of moles NaOH to Fe(NO3)3 in the given solutions is: Ratio = Moles of NaOH / Moles of Fe(NO3)3 = 0.500 mol / 0.150 mol ≈ 3.33 Since this ratio is larger than 2 (the stoichiometric ratio), this means that NaOH is present in excess, and Fe(NO3)3 is the limiting reactant. Using the moles of the limiting reactant, we can find the theoretical yield of precipitate (Fe(OH)3) produced: Moles of Fe(OH)3 formed = Moles of Fe(NO3)3 = 0.150 mol Now, we can convert moles of Fe(OH)3 into grams: Theoretical yield = Moles × Molar mass = 0.150 mol × (56.0 g/mol + (3 × 16.0 g/mol) + (3 × 1.0 g/mol)) = 0.150 mol × 106.0 g/mol = 15.9 g

Step 3: Calculate the percent yield

Given that only 10.95 g of precipitate is isolated, we can now calculate the percent yield: Percent yield = (Actual yield / Theoretical yield) × 100% Percent yield = (10.95 g / 15.9 g) × 100% ≈ 68.87%

Step 4: Find the molar concentration of the excess reactant

From earlier, we concluded that NaOH was the excess reactant. To find the molar concentration of NaOH in the combined solution, we first need to determine the moles of NaOH that actually reacted: Moles of NaOH reacted = 2 × Moles of Fe(NO3)3 = 2 × 0.150 mol = 0.300 mol Since we started with 0.500 mol of NaOH, there are 0.200 mol of NaOH remaining: Moles of NaOH remaining = Initial moles - Moles reacted = 0.500 mol - 0.300 mol = 0.200 mol Finally, we can calculate the molar concentration of NaOH in the 300.0 mL of combined solutions: Molar concentration of NaOH = Moles of NaOH remaining / Total volume Molar concentration = 0.200 mol / (0.200 L + 0.100 L) = 0.200 mol / 0.300 L ≈ 0.667 M

Key Concepts

Theoretical Yield

When combining chemical reactants, it's crucial to predict how much product can be formed. This prediction is known as the theoretical yield, an essential concept in chemistry helping us to understand the maximum amount of product that could be generated from given reactants under ideal conditions. It's calculated based on stoichiometry, which involves the balanced chemical equation of the reaction.

For instance, when sodium hydroxide reacts with iron(III) nitrate to form a precipitate, as shown in our exercise, the theoretical yield informs us of the maximum possible amount of this solid precipitate that can result. By knowing the moles of the limiting reactant (in this case, iron(III) nitrate), students are equipped to calculate the moles of precipitate, and subsequently bring in the molar mass to convert these to grams. This process enables a concrete grasp of the reaction's efficiency before stepping foot in a lab.

Percent Yield

While theoretical yield sets the expectations, actual laboratory results can differ due to various reasons, such as reaction conditions or measurement errors. Percent yield comes into play here, as it quantifies the efficiency of a chemical reaction. This percentage is the ratio of the actual yield (the amount of product actually obtained) to the theoretical yield (the amount that could be formed in an ideal scenario), multiplied by 100%.

Through this calculation, students gain insights on the real-world performance of the reaction. For example, if our exercise concluded with a percent yield of approximately 68.87%, it tells us that the reaction efficiency isn't 100% and helps us ponder on possible causes for this discrepancy. Such understanding not only bolsters theoretical knowledge but also sharpens problem-solving skills, as students learn to account for and troubleshoot experimental inconsistencies.

Molar Concentration

The concept of molar concentration tells us the amount of a substance in a given volume of solution, often measured in moles per liter (M). It's an underpinning idea in the realm of solutions chemistry, guiding predictions about how substances will react when mixed. When you know the concentration of your reactants, you can strategize the quantities necessary for a reaction.

In our exercise, determining the molar concentration of sodium hydroxide post-reaction (having not been consumed entirely) allows for the comprehension of the extent of excess present. The formula used is straightforward: divide the remaining moles of your reactant by the total volume of the solution. By mastering this concept, students can adeptly manipulate reaction quantities, anticipate outcomes, and optimize reactant usage—key skills in any chemist’s arsenal.