General Chemistry Final Exam Study Guide – Step-by-Step Guidance

Q1. Balance the following chemical equations:

• Al4C3(s) + H2O(l) ⟶ Al(OH)3(s) + CH4(g)

• C5H10O2(l) + O2(g) ⟶ CO2(g) + H2O(g)

• Fe(OH)3(s) + H2SO4(aq) ⟶ Fe2(SO4)3(aq) + H2O(l)

• Mg3N2(s) + H2SO4(aq) ⟶ MgSO4(aq) + (NH4)2SO4(aq)

Background

Topic: Chemical Reactions and Stoichiometry

This question tests your ability to balance chemical equations, ensuring the number of atoms of each element is the same on both sides of the equation.

Key Terms and Concepts:

• Reactants: Substances present before the reaction.

• Products: Substances formed as a result of the reaction.

• Balancing: Adjusting coefficients to ensure atom conservation.

Step-by-Step Guidance

1. Write out the unbalanced equation, listing all reactants and products.

2. Count the number of atoms of each element on both sides of the equation.

3. Start balancing with the element that appears in the fewest compounds, usually metals or complex ions.

4. Adjust coefficients (not subscripts) to balance each element, one at a time.

5. Check your work: recount atoms to ensure both sides are equal for every element.

Try solving on your own before revealing the answer!

Final Answer:

• Al4C3(s) + 12 H2O(l) ⟶ 4 Al(OH)3(s) + 3 CH4(g)

• C5H10O2(l) + 7 O2(g) ⟶ 5 CO2(g) + 5 H2O(g)

• 2 Fe(OH)3(s) + 3 H2SO4(aq) ⟶ Fe2(SO4)3(aq) + 6 H2O(l)

• Mg3N2(s) + 4 H2SO4(aq) ⟶ 3 MgSO4(aq) + 2 (NH4)2SO4(aq)

Each equation is balanced so that the number of atoms for each element is equal on both sides.

Q2. Calculate the mass, in grams, of 0.105 mol sucrose (C12H22O11).

Background

Topic: Moles, Mass, and Molar Mass

This question tests your ability to convert between moles and grams using the molar mass of a compound.

Key Terms and Formulas:

• Mole: A unit representing 6.022 × 1023 particles.

• Molar Mass: The mass of one mole of a substance (g/mol).

Key formula:

$\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Step-by-Step Guidance

1. Identify the number of moles given: 0.105 mol.

2. Calculate the molar mass of sucrose (C12H22O11) by summing the atomic masses: $12 \times \text{C} + 22 \times \text{H} + 11 \times \text{O}$.

3. Multiply the number of moles by the molar mass to find the mass in grams.

Try solving on your own before revealing the answer!

Final Answer: 35.9 g

$\text{mass} = 0.105 \times 342.02 = 35.9 \text{ g}$

The molar mass of sucrose is 342.02 g/mol, so multiplying by the number of moles gives the mass.

Q3. Calculate the number of molecules in 1.0 × 10−6 mol CH3CH2OH.

Background

Topic: Molecules, Moles, and Avogadro's Number

This question tests your ability to convert moles to molecules using Avogadro's number.

Key Terms and Formulas:

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

• Mole: A unit for counting particles.

Key formula:

$\text{Number of molecules} = \text{moles} \times 6.022 \times 10^{23}$

Step-by-Step Guidance

1. Identify the number of moles: $1.0 \times 10^{-6}$ mol.

2. Multiply the number of moles by Avogadro's number to find the number of molecules.

Try solving on your own before revealing the answer!

Final Answer: $6.0 \times 10^{17}$ molecules

$1.0 \times 10^{-6} \times 6.022 \times 10^{23} = 6.0 \times 10^{17}$ molecules$

This calculation uses Avogadro's number to convert moles to molecules.

Q4. Calculate the empirical formula of a compound from the following data: 0.0130 mol C, 0.0390 mol H, and 0.0065 mol O.

Background

Topic: Empirical Formula Determination

This question tests your ability to determine the simplest whole-number ratio of atoms in a compound.

Key Terms and Formulas:

• Empirical Formula: The simplest ratio of elements in a compound.

Key steps:

1. List the number of moles for each element: C = 0.0130, H = 0.0390, O = 0.0065.

2. Divide each by the smallest number of moles to get the ratio.

3. If necessary, multiply all ratios by a whole number to get whole numbers for each element.

Try solving on your own before revealing the answer!

Final Answer: C2H6O

Dividing each by 0.0065 gives C:2, H:6, O:1, so the empirical formula is C2H6O.

Q5. Calculate the molarity of a solution that contains 0.175 mol ZnCl2 in exactly 150 mL of solution.

Background

Topic: Solution Concentration (Molarity)

This question tests your ability to calculate molarity, which is moles of solute per liter of solution.

Key Terms and Formulas:

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

Step-by-Step Guidance

1. Identify the number of moles of solute: 0.175 mol ZnCl2.

2. Convert the volume from mL to L: $150 \text{ mL} = 0.150 \text{ L}$.

3. Plug the values into the molarity formula: $\text{M} = \frac{0.175}{0.150}$.

Try solving on your own before revealing the answer!

Final Answer: 1.17 M

$\text{M} = \frac{0.175}{0.150} = 1.17$

This gives the molarity of the ZnCl2 solution.