Chemistry Unit 4 Worksheet 3 Answer Key
The following answer key provides a complete, step‑by‑step solution set for Chemistry Unit 4 Worksheet 3, which focuses on moles, stoichiometry, and limiting reactants. Each problem is broken down into its core concepts, calculations, and final answers, allowing students and teachers to verify work, understand common pitfalls, and reinforce the underlying chemistry principles Simple, but easy to overlook..
Introduction
Unit 4 in most high‑school chemistry curricula introduces the quantitative language of chemistry: moles, molar mass, stoichiometric ratios, and limiting reagents. Worksheet 3 is designed to test students’ ability to convert between mass, moles, and particles, balance chemical equations, and determine theoretical yields. The answer key below follows the worksheet’s original numbering, offering clear explanations that can be used for self‑study or classroom review.
1. Converting Mass to Moles
Problem 1: Convert 12.5 g of NaCl to moles.
- Molar mass of NaCl = 22.99 (g mol⁻¹) + 35.45 (g mol⁻¹) = 58.44 g mol⁻¹.
- Moles = mass ÷ molar mass = 12.5 g ÷ 58.44 g mol⁻¹ = 0.214 mol (3 sf).
Answer: 0.214 mol NaCl Turns out it matters..
Problem 2: How many molecules are present in 0.350 mol of CO₂?
- Avogadro’s number = 6.022 × 10²³ molecules mol⁻¹.
- Molecules = 0.350 mol × 6.022 × 10²³ = 2.11 × 10²³ molecules.
Answer: 2.11 × 10²³ CO₂ molecules Turns out it matters..
2. Balancing Chemical Equations
Problem 3: Balance the combustion of propane:
[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} ]
Balanced equation:
[ \boxed{\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}} ]
Explanation: 3 C atoms → 3 CO₂, 8 H atoms → 4 H₂O (needs 8 H), O atoms on right = 3 × 2 + 4 × 1 = 10, therefore 5 O₂ on left Small thing, real impact..
Problem 4: Balance the reaction between aluminum and iron(III) oxide:
[ \text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} ]
Balanced equation:
[ \boxed{2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}} ]
3. Stoichiometric Calculations
3.1. Mass‑to‑Mass Conversions
Problem 5: How many grams of H₂O are produced when 5.00 g of H₂ reacts with excess O₂?
- Balanced equation: 2 H₂ + O₂ → 2 H₂O.
- Moles of H₂: 5.00 g ÷ 2.016 g mol⁻¹ = 2.48 mol H₂.
- Mole ratio (H₂ → H₂O): 1 : 1 (because 2 H₂ → 2 H₂O).
- Moles of H₂O formed: 2.48 mol.
- Mass of H₂O: 2.48 mol × 18.015 g mol⁻¹ = 44.7 g (3 sf).
Answer: 44.7 g H₂O.
Problem 6: 12.0 g of CaCO₃ is heated to produce CaO and CO₂. How many grams of CO₂ are released?
- Balanced equation: CaCO₃ → CaO + CO₂.
- Moles CaCO₃: 12.0 g ÷ 100.09 g mol⁻¹ = 0.1199 mol.
- Mole ratio CaCO₃ → CO₂: 1 : 1, so 0.1199 mol CO₂.
- Mass CO₂: 0.1199 mol × 44.01 g mol⁻¹ = 5.28 g (3 sf).
Answer: 5.28 g CO₂.
3.2. Limiting Reactant & Theoretical Yield
Problem 7: In the reaction
[ \text{2Al} + \text{3Cl}_2 \rightarrow \text{2AlCl}_3 ]
2.00 g of Al reacts with 5.00 g of Cl₂. Determine the limiting reactant and calculate the mass of AlCl₃ formed.
- Molar masses: Al = 26.98 g mol⁻¹; Cl₂ = 70.90 g mol⁻¹; AlCl₃ = 133.34 g mol⁻¹.
- Moles Al: 2.00 g ÷ 26.98 g mol⁻¹ = 0.0741 mol.
- Moles Cl₂: 5.00 g ÷ 70.90 g mol⁻¹ = 0.0705 mol.
- Stoichiometric requirement: 2 mol Al per 3 mol Cl₂ → ratio Al/Cl₂ = 2/3 = 0.667.
- Required Cl₂ for 0.0741 mol Al = 0.0741 mol × (3/2) = 0.111 mol, but only 0.0705 mol is present.
- Cl₂ is limiting.
- Moles AlCl₃ formed: From Cl₂, 3 mol Cl₂ → 2 mol AlCl₃, so 0.0705 mol × (2/3) = 0.0470 mol AlCl₃.
- Mass AlCl₃: 0.0470 mol × 133.34 g mol⁻¹ = 6.27 g (3 sf).
Answer: Cl₂ limits the reaction; 6.27 g AlCl₃ is produced Easy to understand, harder to ignore..
Problem 8: A mixture contains 10.0 g of Na₂CO₃ and 8.0 g of HCl. Reaction:
[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 ]
Find the limiting reactant and the mass of NaCl formed.
- Molar masses: Na₂CO₃ = 105.99 g mol⁻¹; HCl = 36.46 g mol⁻¹; NaCl = 58.44 g mol⁻¹.
- Moles Na₂CO₃: 10.0 g ÷ 105.99 g mol⁻¹ = 0.0944 mol.
- Moles HCl: 8.0 g ÷ 36.46 g mol⁻¹ = 0.219 mol.
- Stoichiometry: 1 mol Na₂CO₃ requires 2 mol HCl. Required HCl = 0.0944 mol × 2 = 0.1888 mol, which is less than the 0.219 mol available.
- Na₂CO₃ is limiting.
- Moles NaCl produced: 1 mol Na₂CO₃ → 2 mol NaCl, so 0.0944 mol × 2 = 0.1888 mol NaCl.
- Mass NaCl: 0.1888 mol × 58.44 g mol⁻¹ = 11.0 g (3 sf).
Answer: Na₂CO₃ limits the reaction; 11.0 g NaCl formed.
4. Percent Yield
Problem 9: In a laboratory synthesis of ammonia, the theoretical yield is 25.0 g, but only 18.5 g are actually recovered. Calculate the percent yield That's the part that actually makes a difference..
[ % \text{Yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{18.5}{25.0}\times100 = \boxed{74%} ]
Answer: 74 % yield.
Problem 10: An experiment produces 4.20 g of CuSO₄·5H₂O, while the calculated theoretical yield is 5.00 g. What is the percent yield?
[ % \text{Yield} = \frac{4.20}{5.00}\times100 = 84% ]
Answer: 84 % yield.
5. Empirical and Molecular Formulas
Problem 11: A compound contains 40.0 % C, 6.7 % H, and 53.3 % O by mass. Determine its empirical formula Small thing, real impact..
- Assume 100 g sample: C = 40.0 g, H = 6.7 g, O = 53.3 g.
- Convert to moles:
- C: 40.0 g ÷ 12.01 g mol⁻¹ = 3.33 mol
- H: 6.7 g ÷ 1.008 g mol⁻¹ = 6.65 mol
- O: 53.3 g ÷ 16.00 g mol⁻¹ = 3.33 mol
- Divide by smallest (3.33):
- C = 1, H ≈ 2, O = 1.
Empirical formula: CH₂O.
Problem 12: The molar mass of the compound in Problem 11 is 180 g mol⁻¹. Find its molecular formula Not complicated — just consistent..
- Empirical formula mass (CH₂O) = 12.01 + 2 × 1.008 + 16.00 = 30.03 g mol⁻¹.
- Multiplication factor = 180 ÷ 30.03 ≈ 6.
- Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
Answer: C₆H₁₂O₆ (glucose).
6. Gas‑Law Applications (Ideal Gas Approximation)
Problem 13: Calculate the volume (at STP) of 0.250 mol of N₂ gas.
- At STP, 1 mol gas = 22.4 L.
- Volume = 0.250 mol × 22.4 L mol⁻¹ = 5.60 L.
Answer: 5.60 L N₂ Worth keeping that in mind..
Problem 14: A sample of CO₂ occupies 12.5 L at 298 K and 1.00 atm. What is the amount in moles?
Using PV = nRT (R = 0.0821 L·atm mol⁻¹·K⁻¹):
[ n = \frac{PV}{RT} = \frac{1.00 \times 12.In real terms, 5}{0. Even so, 0821 \times 298} = \frac{12. 5}{24.5} = 0 And it works..
Answer: 0.511 mol CO₂.
7. Common Mistakes & Tips
| Mistake | Why it Happens | How to Avoid |
|---|---|---|
| Forgetting to balance the equation before stoichiometry | Students jump straight to mass‑to‑mass calculations. Consider this: | Always write and balance the equation first; the coefficients become the conversion factors. |
| Using atomic mass instead of molar mass | Confusing the unit (g mol⁻¹) with a simple atomic weight. | Remember that molar mass = atomic mass × Avogadro’s number, expressed in g mol⁻¹. |
| Mixing up limiting reactant vs. excess reactant | Overlooking the stoichiometric ratio. | Compare the available mole ratio to the required ratio; the smaller quotient indicates the limiting species. Which means |
| Rounding too early | Early rounding propagates error through multiple steps. Practically speaking, | Keep at least 4–5 significant figures during calculations; round only for the final answer. |
| Ignoring temperature/pressure corrections for gases | Assuming all gases behave ideally at any condition. | Use the Ideal Gas Law and convert to standard conditions when required. |
8. Frequently Asked Questions
Q1. Why do we use 22.4 L for a gas at STP?
A: At Standard Temperature and Pressure (0 °C, 1 atm), one mole of any ideal gas occupies 22.4 L. This value derives from the Ideal Gas Law (PV = nRT) with the defined constants.
Q2. Can the limiting reactant be determined by mass alone?
A: Only if the masses are converted to moles and compared using the balanced equation. Direct mass comparison without mole conversion can be misleading because different substances have different molar masses Small thing, real impact..
Q3. What is the difference between theoretical yield and percent yield?
A: The theoretical yield is the maximum amount of product predicted by stoichiometry, assuming 100 % efficiency. Percent yield quantifies the actual experimental result relative to that ideal amount.
Q4. How do I know if a compound’s empirical formula needs to be multiplied?
A: Compare the empirical formula mass to the given molar mass. If the molar mass is an integer multiple of the empirical mass, multiply the subscripts by that factor to obtain the molecular formula Still holds up..
Q5. When is it appropriate to use the mole‑ratio method vs. the mass‑ratio method?
A: Both lead to the same answer, but the mole‑ratio method is generally clearer because it follows directly from the balanced equation. Use the mass‑ratio method only when you’re comfortable converting masses to moles implicitly.
9. Conclusion
The answer key for Chemistry Unit 4 Worksheet 3 not only supplies the correct numerical results but also illustrates the logical flow behind each calculation. Even so, mastery of mole concepts, balanced equations, limiting‑reactant analysis, and percent‑yield determinations equips students to tackle more complex quantitative problems in later units, such as solution stoichiometry and thermochemistry. By reviewing the step‑by‑step solutions, students can identify where they may have gone astray, reinforce the fundamental principles, and build confidence for future laboratory work and examinations.
Tip for teachers: Use the detailed explanations as a scaffold for in‑class discussions. Prompt learners to explain each step in their own words; this active recall deepens understanding and improves retention far beyond simple answer checking.
