General Chemistry Exam Study Guidance

Q1. How many aluminum ions are present in 65.5 mL of 0.210 M AlCl3 solution?

Background

Topic: Solution Stoichiometry & Ion Counting

This question tests your ability to use molarity and volume to determine the number of ions in a solution.

Key Terms and Formulas

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Avogadro's Number: $6.022 \times 10^{23}$ particles/mol

• AlCl3 dissociation: $\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^-$

Step-by-Step Guidance

1. Convert the volume from mL to L: $65.5\ \text{mL} = \frac{65.5}{1000}\ \text{L}$

2. Calculate the moles of AlCl3 using molarity: $\text{moles AlCl}_3 = M \times \text{volume (L)}$

3. Determine the moles of Al3+ ions produced per mole of AlCl3 (1:1 ratio).

4. Calculate the number of Al3+ ions using Avogadro's number: $\text{number of ions} = \text{moles Al}^{3+} \times 6.022 \times 10^{23}$

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Q2. If the percent yield for the following reaction is 75.0%, and 25.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced?

3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

Background

Topic: Stoichiometry & Percent Yield

This question tests your ability to use stoichiometry and percent yield to determine the actual amount of product formed.

Key Terms and Formulas

• Percent Yield: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Molar Masses: Find the molar masses of NO2 and HNO3.

Step-by-Step Guidance

1. Convert 25.0 g NO2 to moles: $\text{moles NO}_2 = \frac{25.0\ \text{g}}{\text{molar mass of NO}_2}$

2. Use the stoichiometric ratio from the balanced equation to find moles of HNO3 produced (3:2 ratio).

3. Convert moles of HNO3 to grams: $\text{grams HNO}_3 = \text{moles HNO}_3 \times \text{molar mass of HNO}_3$

4. Apply percent yield to find the actual yield: $\text{Actual Yield} = \text{Theoretical Yield} \times \frac{75.0}{100}$

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Q3. How many grams of NaCl are required to make 300.0 mL of a 2.500 M solution?

Background

Topic: Solution Preparation

This question tests your ability to calculate the mass of solute needed to prepare a solution of a given molarity and volume.

Key Terms and Formulas

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Moles to grams: $\text{mass} = \text{moles} \times \text{molar mass}$

Step-by-Step Guidance

1. Convert 300.0 mL to liters: $\frac{300.0}{1000} = 0.3000\ \text{L}$

2. Calculate moles of NaCl needed: $\text{moles NaCl} = M \times \text{volume (L)}$

3. Find the molar mass of NaCl (Na: 22.99, Cl: 35.45 g/mol).

4. Calculate the mass of NaCl: $\text{mass} = \text{moles} \times \text{molar mass}$

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Q4. If the percent yield for the following reaction is 65.0%, how many grams of KClO3 are needed to produce 8.00 g of O2?

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Background

Topic: Stoichiometry & Percent Yield

This question tests your ability to use stoichiometry and percent yield to determine the amount of reactant needed to produce a certain amount of product.

Key Terms and Formulas

• Percent Yield: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Molar Masses: Find the molar masses of O2 and KClO3.

Step-by-Step Guidance

1. Convert 8.00 g O2 to moles: $\text{moles O}_2 = \frac{8.00\ \text{g}}{\text{molar mass of O}_2}$

2. Use the percent yield to find the theoretical yield: $\text{Theoretical Yield} = \frac{\text{Actual Yield}}{\text{Percent Yield}} \times 100\%$

3. Use the stoichiometric ratio from the balanced equation to find moles of KClO3 needed (3:2 ratio with O2).

4. Convert moles of KClO3 to grams: $\text{mass} = \text{moles} \times \text{molar mass of KClO}_3$

Try solving on your own before revealing the answer!

Q5. The mixing of which pair of reactants will result in a precipitation reaction?

Background

Topic: Precipitation Reactions & Solubility Rules

This question tests your knowledge of solubility rules and your ability to predict whether a precipitate will form when two solutions are mixed.

Key Terms and Formulas

• Precipitation Reaction: A reaction in which an insoluble solid (precipitate) forms when two solutions are mixed.

• Solubility Rules: Guidelines for predicting the solubility of ionic compounds in water.

Step-by-Step Guidance

1. Write the possible products for each pair by exchanging ions (double displacement).

2. Use solubility rules to determine if any of the products are insoluble in water.

3. Identify which pair produces an insoluble compound (the precipitate).

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Q6. How many molecules of HCl are formed when 60.0 g of water reacts according to the following balanced reaction? Assume excess ICl3. (Answer must be in scientific notation)

2 ICl3 + 3 H2O → ICl + HIO3 + 5 HCl

Background

Topic: Stoichiometry & Molecule Counting

This question tests your ability to use stoichiometry to relate mass of a reactant to the number of molecules of a product.

Key Terms and Formulas

• Molar Mass of H2O: 18.02 g/mol

• Stoichiometric Ratios: From the balanced equation, 3 mol H2O produces 5 mol HCl.

• Avogadro's Number: $6.022 \times 10^{23}$ molecules/mol

Step-by-Step Guidance

1. Convert 60.0 g H2O to moles: $\text{moles H}_2\text{O} = \frac{60.0\ \text{g}}{18.02\ \text{g/mol}}$

2. Use the stoichiometric ratio to find moles of HCl produced: $\text{moles HCl} = \text{moles H}_2\text{O} \times \frac{5}{3}$

3. Convert moles of HCl to molecules: $\text{molecules HCl} = \text{moles HCl} \times 6.022 \times 10^{23}$

Try solving on your own before revealing the answer!