Stoichiometry Murder Mystery Answer Key

Stoichiometry Murder Mystery: The Case of the Missing Magnesium – Answer Key

This document provides the answers and detailed explanations for the "Stoichiometry Murder Mystery" case. Think about it: this engaging activity uses stoichiometric calculations to solve a fictional crime, helping students solidify their understanding of mole ratios, limiting reactants, and percent yield. So understanding these concepts is crucial for success in chemistry and related fields. This answer key is designed to be used by educators to guide students, offering a step-by-step solution for each part of the mystery.

The Case: A renowned chemist, Dr. Albert Einstein (fictional, of course!), was found dead in his lab. The only clue: a partially reacted magnesium ribbon and a note mentioning a specific reaction. The task is to use the available evidence to determine what happened and identify the culprit.

Part 1: The Reaction

The note left by Dr. Einstein described the reaction:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This is a classic single-displacement reaction where magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas.

Question 1: Balancing the Equation

The equation is already balanced. One magnesium atom, two hydrogen atoms, and two chlorine atoms are present on both the reactant and product sides.

Part 2: The Evidence

The investigators found the following evidence:

• 2.43 grams of magnesium ribbon: This represents the initial mass of the magnesium used in the reaction.
• 50.0 mL of 1.0 M HCl: This is the volume and concentration of the hydrochloric acid used.
• A flask containing 100 mL of hydrogen gas collected over water at 25°C and 1 atm: This represents the product of the reaction. (The water vapor pressure at 25°C needs to be considered). The pressure of dry hydrogen gas is needed.

Part 3: Calculations and Deductions

This section uses stoichiometry to analyse the evidence and solve the mystery.

Question 2: Moles of Magnesium

First, we need to calculate the moles of magnesium used:

• Molar mass of Mg = 24.31 g/mol
• Moles of Mg = (mass of Mg) / (molar mass of Mg) = 2.43 g / 24.31 g/mol = 0.100 mol

Question 3: Moles of HCl

Next, calculate the moles of HCl used:

• Moles of HCl = (volume of HCl in liters) x (concentration of HCl) = (0.050 L) x (1.0 mol/L) = 0.050 mol

Question 4: Limiting Reactant

Determine the limiting reactant by comparing the mole ratio of magnesium to HCl in the balanced equation (1:2).

• For Mg: 0.100 mol Mg x (2 mol HCl / 1 mol Mg) = 0.200 mol HCl required.
• Since only 0.050 mol HCl was present, HCl is the limiting reactant.

Question 5: Theoretical Yield of H₂

Calculate the theoretical yield of hydrogen gas (H₂) using the moles of the limiting reactant (HCl):

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• From the balanced equation: 2 mol HCl produces 1 mol H₂
• Moles of H₂ = (moles of HCl) x (1 mol H₂ / 2 mol HCl) = 0.050 mol HCl x (1 mol H₂ / 2 mol HCl) = 0.025 mol H₂

Question 6: Volume of Dry H₂ at STP

Convert the moles of H₂ to volume at Standard Temperature and Pressure (STP, 0°C and 1 atm), ignoring the water vapor for the moment:

• At STP, 1 mol of any gas occupies 22.4 L.
• Volume of H₂ at STP = (moles of H₂ ) x (22.4 L/mol) = 0.025 mol x 22.4 L/mol = 0.56 L

Question 7: Correcting for Water Vapor Pressure

The hydrogen gas was collected over water; therefore, the collected gas is a mixture of hydrogen and water vapor. We must correct for the partial pressure of water vapor at 25°C (look up the vapor pressure in a reference table; approximately 23.8 mmHg).

• Water vapor pressure = 23.8 mmHg / 760 mmHg/atm = 0.0313 atm
• Total pressure = 1 atm
• Partial pressure of H₂ = Total pressure - Water vapor pressure = 1 atm - 0.0313 atm = 0.9687 atm

Now, use the ideal gas law (PV = nRT) to calculate the volume of dry H₂ at 25°C and 0.9687 atm. In real terms, r = 0. 0821 L·atm/mol·K; T = 25°C + 273.15 = 298.

• V = nRT/P = (0.025 mol)(0.0821 L·atm/mol·K)(298.15 K) / 0.9687 atm ≈ 0.63 L

Question 8: Percent Yield

Compare the actual yield (0.63 L) to the theoretical yield at STP (0.56L) to determine the percent yield.

• Percent yield = (Actual yield / Theoretical yield) x 100% = (0.63 L / 0.56 L) x 100% ≈ 112.5%

Part 4: The Solution to the Mystery

The percent yield is significantly greater than 100%. This indicates that something else was producing hydrogen gas in addition to the reaction between magnesium and hydrochloric acid. This extra hydrogen gas suggests that another reactant that produces hydrogen was present.

Question 9: The Culprit

The most likely scenario is that another substance was added to the reaction flask to produce extra hydrogen gas, possibly in a deliberate attempt to harm Dr. On top of that, einstein. The high percent yield and the presence of unreacted magnesium point to a hidden source of hydrogen production. In practice, this was likely a deliberate act of sabotage. Consider this: the culprit used a substance which also reacted with the HCl to produce additional hydrogen gas. Without more forensic evidence, the exact substance cannot be determined definitively.

Part 5: Further Investigations (Optional)

This part encourages students to explore the possibilities and formulate hypotheses. This could involve researching other chemicals that react with HCl to produce hydrogen gas.

Conclusion:

This stoichiometry murder mystery provides a stimulating and interactive way for students to practice their stoichiometric calculations while developing their problem-solving skills. The unexpected high percent yield serves as a crucial clue, emphasizing the importance of error analysis and critical thinking in scientific investigations. The activity also highlights the practical uses of stoichiometry in forensic science and investigative work. In real terms, the open-ended nature of the final question encourages further exploration and discussion, reinforcing the application of chemical principles in a real-world context. By carefully analyzing the experimental data and applying stoichiometric principles, students can solve the mystery and gain a deeper understanding of the concepts involved.