Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Define and calculate molarity of solutions.
• Perform stoichiometric calculations for reactions in solution.
• Apply the ideal gas law (PV = nRT).
• Perform stoichiometric calculations involving gases.
• Determine the percentage yield of a reaction.

Lesson summary

Chemical reactions are the backbone of industry, medicine, and even the natural world around us. Stoichiometry is the mathematical language we use to describe these reactions, allowing us to predict how much of each reactant we need and how much product we will get. This week, we'll focus on applying stoichiometric principles to solutions and gases, which are particularly important in many industrial processes in South Africa, such as mining and chemical manufacturing. For example, understanding the stoichiometry of acid mine drainage is crucial for managing its environmental impact.

Lesson notes

2.1 Solutions and Molarity 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 this section, we'll focus on aqueous solutions (where water is the solvent). Molarity (M) is a measure of the concentration of a solution. It is defined as the number of moles of solute per liter of solution.

The formula is: Molarity (M) = (moles of solute) / (volume of solution in liters) M = n/V Example 1: What is the molarity of a solution prepared by dissolving 10.0 g of sodium chloride (NaCl) in enough water to make 500.0 mL of solution?

Step 1: Convert grams of NaCl to moles. The molar mass of NaCl is 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol. Moles of NaCl = (10.0 g) / (58.44 g/mol) = 0.171 mol Step 2: Convert milliliters to liters. Volume of solution = 500.0 mL = 0.5000 L Step 3: Calculate the molarity. Molarity = (0.171 mol) / (0.5000 L) = 0.342 M Therefore, the molarity of the NaCl solution is 0.342

M. Example 2: How many grams of potassium hydroxide (KOH) are needed to prepare 250.0 mL of a 0.200 M solution?

Step 1: Convert milliliters to liters. Volume of solution = 250.0 mL = 0.2500 L Step 2: Calculate the number of moles of KOH needed. Moles of KOH = (Molarity) x (Volume) = (0.200 M) x (0.2500 L) = 0.0500 mol Step 3: Convert moles of KOH to grams. The molar mass of KOH is 39.10 g/mol (K) + 16.00 g/mol (O) + 1.01 g/mol (H) = 56.11 g/mol. Grams of KOH = (0.0500 mol) x (56.11 g/mol) = 2.81 g Therefore, 2.81 g of KOH are needed to prepare 250.0 mL of a 0.200 M solution. 2.2 Stoichiometry in Solutions When reactions occur in solution, we use molarity to relate the volume of a solution to the number of moles of solute.

Let's look at an example: Example 3: What volume of 0.100 M hydrochloric acid (HCl) is required to completely neutralize 25.0 mL of 0.200 M sodium hydroxide (NaOH)?

Step 1: Write the balanced chemical equation. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) The equation is already balanced, with a 1:1 mole ratio between HCl and NaO

H. Step 2: Calculate the moles of NaO

H. Moles of NaOH = (Molarity) x (Volume) = (0.200 M) x (0.0250 L) = 0.00500 mol Step 3: Determine the moles of HCl required.

Since the mole ratio is 1:1, 0.00500 mol of HCl is required.

Step 4: Calculate the volume of 0.100 M HCl needed. Volume of HCl = (Moles) / (Molarity) = (0.00500 mol) / (0.100 M) = 0.0500 L = 50.0 mL Therefore, 50.0 mL of 0.100 M HCl is required to neutralize the NaOH solution. 2.3 The Ideal Gas Law The ideal gas law relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas: PV = nRT where R is the ideal gas constant. The value of R depends on the units used for pressure and volume. A commonly used value is R = 8.314 J/(mol·K) if pressure is in Pascals (Pa) and volume is in m 3 , or R = 0.0821 L·atm/(mol·K) if pressure is in atm and volume is in

L. Example 4: A gas occupies a volume of 10.0 L at a temperature of 27°C and a pressure of 150 kPa. How many moles of gas are present?

Step 1: Convert temperature to Kelvin. T(K) = T(°C) + 273.15 = 27 + 273.15 = 300.15 K Step 2: Choose an appropriate value for R. Since Pressure is in kPa and we want to find 'n' (moles), using the value R = 8.314 J/(molK) will work, provided we convert Volume to m 3 and Pressure to Pa. Alternatively, we could convert Pressure to atm and Volume to Liters and use R = 0.0821 Latm/(molK). We'll choose the first option for this example.

Step 3: Convert volume to m 3 and pressure to Pa Volume = 10 L = 0.01 m 3 Pressure = 150 kPa = 150000 Pa Step 4: Rearrange the ideal gas law to solve for n. n = PV / RT = (150000 Pa 0.01 m 3 ) / (8.314 J/(mol·K) * 300.15 K) = 0.60 mol Therefore, there are approximately 0.60 moles of gas present. 2.4 Stoichiometry with Gases We can use the ideal gas law to convert between volume and moles of gases in chemical reactions. At Standard Temperature and Pressure (STP), which is defined as 0°C (273.15 K) and 1 atm (101.325 kPa), one mole of any ideal gas occupies a volume of 22.4

L. This is known as the molar volume of a gas at ST

P. Example 5: What volume of oxygen gas (O2) at STP is required to completely react with 5.0 g of magnesium (Mg) according to the following reaction? 2Mg(s) + O2(g) → 2MgO(s)

Step 1: Convert grams of Mg to moles. The molar mass of Mg is 24.31 g/mol. Moles of Mg = (5.0 g) / (24.31 g/mol) = 0.206 mol Step 2: Determine the moles of O2 required. From the balanced equation, the mole ratio of Mg to O2 is 2:

1. Moles of O2 = (0.206 mol Mg) / 2 = 0.103 mol Step 3: Calculate the volume of O2 at STP. Volume of O2 = (Moles) x (Molar Volume at STP) = (0.103 mol) x (22.4 L/mol) = 2.31 L Therefore, 2.31 L of oxygen gas at STP is required to completely react with 5.0 g of magnesium.

Example 6: Ammonia gas (NH3) reacts with oxygen gas (O2) to produce nitrogen monoxide gas (NO) and water vapor (H2O).