Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Define and apply molarity in calculations.
• Use the ideal gas law (pV = nRT) in stoichiometry.
• Find the limiting reactant and theoretical yield.
• Calculate the percentage yield of a reaction.
• Solve practical problems with solutions and gases.

Lesson summary

This week, we delve into the quantitative relationships between reactants and products in chemical reactions, specifically focusing on stoichiometry involving solutions and gases. Stoichiometry is fundamental to understanding and predicting the outcomes of chemical reactions. It allows us to calculate the amounts of reactants needed or products formed in chemical processes. This knowledge is crucial in various fields, from industrial chemical production (e.g., fertilizer production, pharmaceutical manufacturing) to environmental monitoring (e.g., calculating pollutant concentrations) and even in everyday cooking and baking.

Lesson notes

2. 1. Solutions and Molarity (Concentration) A solution is a homogeneous mixture of two or more substances. The solute is the substance being dissolved, and the solvent is the substance doing the dissolving. In aqueous solutions, water is the solvent.

Molarity (M): Molarity is a measure of the concentration of a solution, defined as the number of moles of solute per liter of solution.

Formula: Molarity (M) = moles of solute (n) / volume of solution in liters (V)

Units: mol/L or M Example 1: Calculate the molarity of a solution prepared by dissolving 10.0 g of sodium chloride (NaCl) in enough water to make 500 mL of solution.

Step 1: Calculate the number of moles of NaCl. Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol Moles of NaCl (n) = mass / molar mass = 10.0 g / 58.44 g/mol = 0.171 mol Step 2: Convert the volume of the solution to liters. Volume (V) = 500 mL = 0.500 L Step 3: Calculate the molarity. Molarity (M) = n / V = 0.171 mol / 0.500 L = 0.342 M Therefore, the molarity of the NaCl solution is 0.342

M. Example 2: What mass of potassium hydroxide (KOH) is required to prepare 250 mL of a 0.200 M solution?

Step 1: Calculate the number of moles of KOH needed. Moles of KOH (n) = Molarity (M) × Volume (V) = 0.200 mol/L × 0.250 L = 0.0500 mol Step 2: Calculate the mass of KOH needed. Molar mass of KOH = 39.10 g/mol (K) + 16.00 g/mol (O) + 1.01 g/mol (H) = 56.11 g/mol Mass of KOH = moles (n) × molar mass = 0.0500 mol × 56.11 g/mol = 2.81 g Therefore, 2.81 g of KOH is required. 2.

2. Gases and the Ideal Gas Law The Ideal Gas Law describes the relationship between pressure (p), volume (V), number of moles (n), and temperature (T) for an ideal gas. While no gas is truly "ideal," many gases behave closely enough to ideal behavior under typical conditions, making the Ideal Gas Law a useful approximation.

Ideal Gas Law: pV = nRT p = pressure (in Pascals, Pa, if using R = 8.314 J/mol·K, or in atmospheres, atm, if using R = 0.0821 L·atm/mol·K) V = volume (in cubic meters, m³, if using R = 8.314 J/mol·K, or in liters, L, if using R = 0.0821 L·atm/mol·K) n = number of moles R = Ideal Gas Constant (8.314 J/mol·K or 0.0821 L·atm/mol·K) T = temperature (in Kelvin, K) (K = °C + 273.15)

Example 3: What volume will 2.00 moles of oxygen gas (O₂) occupy at a pressure of 101.3 kPa and a temperature of 25 °C?

Step 1: Convert units to be consistent with the value of R (e.g., R=8.314 J/mol·K). Pressure (p) = 101.3 kPa = 101300 Pa Temperature (T) = 25 °C = 25 + 273.15 = 298.15 K Step 2: Rearrange the Ideal Gas Law to solve for volume (V). V = nRT / p Step 3: Substitute the values and calculate

V. V = (2.00 mol × 8.314 J/mol·K × 298.15 K) / 101300 Pa = 0.0491 m³ Step 4: Convert m³ to L (1 m³ = 1000 L) V = 0.0491 m³ 1000 L/m³ = 49.1 L Therefore, the volume of oxygen gas is 49.1 L. 2.

3. Limiting Reactant, Theoretical Yield, and Percentage Yield Limiting Reactant: In a chemical reaction with multiple reactants, the limiting reactant is the reactant that is completely consumed first. It determines the maximum amount of product that can be formed. The other reactants are said to be in excess.

Theoretical Yield: The theoretical yield is the maximum amount of product that can be produced from a given amount of limiting reactant, assuming the reaction goes to completion (100% efficiency).

Actual Yield: The actual yield is the amount of product actually obtained from a reaction. It is usually less than the theoretical yield due to factors like incomplete reactions, side reactions, and loss of product during isolation and purification.

Percentage Yield: The percentage yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

Formula: Percentage Yield = (Actual Yield / Theoretical Yield) × 100% Example 4: Consider the reaction: 2H₂(g) + O₂(g) → 2H₂O(g) If 4.0 g of H₂ reacts with 32.0 g of O₂, what is the limiting reactant, the theoretical yield of H₂O (in grams), and if 30.0 g of water are actually produced, the percentage yield?

Step 1: Calculate the moles of each reactant. Moles of H₂ = 4.0 g / 2.02 g/mol = 1.98 mol Moles of O₂ = 32.0 g / 32.00 g/mol = 1.00 mol Step 2: Determine the limiting reactant. From the balanced equation, 2 moles of H₂ react with 1 mole of O₂. To react with 1.00 mol of O₂, you need 2.00 mol of H₂. Since we only have 1.98 mol of H₂, H₂ is the limiting reactant.

Step 3: Calculate the theoretical yield of H₂O. From the balanced equation, 2 moles of H₂ produce 2 moles of H₂O.

Therefore, 1.98 mol of H₂ will produce 1.98 mol of H₂

O. Molar mass of H₂O = 18.02 g/mol Theoretical yield of H₂O = 1.98 mol × 18.02 g/mol = 35.7 g Step 4: Calculate the percentage yield. Percentage Yield = (Actual Yield / Theoretical Yield) × 100% = (30.0 g / 35.7 g) × 100% = 84.0% Therefore, the limiting reactant is H₂, the theoretical yield of H₂O is 35.7 g, and the percentage yield is 84.0%.