Stoichiometry Worksheet 1 Mass Mass Answer Key

Stoichiometry Worksheet 1: Mass-Mass Stoichiometry Problems – A Comprehensive Guide with Answers

Stoichiometry, at its core, is the study of the quantitative relationships between reactants and products in chemical reactions. Mastering stoichiometry is crucial for any aspiring chemist, allowing you to predict the amount of product formed from a given amount of reactant, or vice versa. This article serves as a comprehensive guide to mass-mass stoichiometry problems, providing a detailed explanation, worked examples, and solutions to a sample worksheet. We'll tackle various complexities to ensure a solid understanding of this fundamental concept.

Understanding Mass-Mass Stoichiometry

Mass-mass stoichiometry problems involve calculating the mass of a product formed or reactant consumed given the mass of another reactant or product. These problems hinge on the mole concept and the balanced chemical equation. The key steps involved are:

1. Balancing the Chemical Equation: Ensure the equation accurately reflects the law of conservation of mass, where the number of atoms of each element is equal on both sides of the equation.

2. Converting Grams to Moles: Using the molar mass of the substance (found on the periodic table), convert the given mass of the reactant or product into moles. Remember, molar mass is the mass of one mole of a substance in grams.

3. Using Mole Ratios: The coefficients in the balanced chemical equation provide the mole ratio between reactants and products. This ratio is crucial for determining the number of moles of the desired substance.

4. Converting Moles to Grams: Finally, convert the moles of the desired substance back into grams using its molar mass.

Example Problem 1: Combustion of Methane

Let's consider the combustion of methane (CH₄):

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Problem: If 16 grams of methane (CH₄) are completely combusted, what mass of carbon dioxide (CO₂) is produced?

Solution:

1. Molar Mass:

   • CH₄: 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol
   • CO₂: 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol

2. Moles of CH₄:

   • Moles = mass / molar mass = 16 g / 16.05 g/mol ≈ 0.997 moles

3. Mole Ratio: From the balanced equation, the mole ratio of CH₄ to CO₂ is 1:1. Therefore, 0.997 moles of CH₄ will produce 0.997 moles of CO₂.

4. Mass of CO₂:

   • Mass = moles * molar mass = 0.997 moles * 44.01 g/mol ≈ 43.87 g

Answer: Approximately 43.87 grams of carbon dioxide (CO₂) are produced.

Example Problem 2: Reaction of Sodium with Chlorine

Consider the reaction between sodium (Na) and chlorine (Cl₂) to form sodium chloride (NaCl):

2Na(s) + Cl₂(g) → 2NaCl(s)

Problem: If 46 grams of sodium react completely with chlorine gas, what mass of sodium chloride is formed?

Solution:

1. Molar Mass:

   • Na: 22.99 g/mol
   • Cl₂: 2 * 35.45 g/mol = 70.90 g/mol
   • NaCl: 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol

2. Moles of Na:

   • Moles = mass / molar mass = 46 g / 22.99 g/mol ≈ 2.00 moles

3. Mole Ratio: The mole ratio of Na to NaCl is 2:2, which simplifies to 1:1. Therefore, 2.00 moles of Na will produce 2.00 moles of NaCl.

4. Mass of NaCl:

   • Mass = moles * molar mass = 2.00 moles * 58.44 g/mol = 116.88 g

Answer: 116.88 grams of sodium chloride (NaCl) are formed.

Example Problem 3: Limiting Reactant Problems

Mass-mass stoichiometry often involves limiting reactants. The limiting reactant is the reactant that is completely consumed first, thereby limiting the amount of product formed. Let's illustrate this:

Problem: 20 grams of hydrogen gas (H₂) reacts with 100 grams of oxygen gas (O₂) to produce water (H₂O). What mass of water is produced?

2H₂(g) + O₂(g) → 2H₂O(l)

Solution:

1. Molar Mass:

   • H₂: 2 * 1.01 g/mol = 2.02 g/mol
   • O₂: 2 * 16.00 g/mol = 32.00 g/mol
   • H₂O: 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol

2. Moles of Reactants:

   • Moles of H₂ = 20 g / 2.02 g/mol ≈ 9.90 moles
   • Moles of O₂ = 100 g / 32.00 g/mol ≈ 3.13 moles

3. Determining the Limiting Reactant: From the balanced equation, the mole ratio of H₂ to O₂ is 2:1. For every 1 mole of O₂, you need 2 moles of H₂. Since we have 3.13 moles of O₂, we'd need 2 * 3.13 = 6.26 moles of H₂. We have more H₂ than needed (9.90 moles). Therefore, O₂ is the limiting reactant.

4. Moles of H₂O: Using the mole ratio from the balanced equation (1 mole O₂ produces 2 moles H₂O), we have:

   • Moles of H₂O = 3.13 moles O₂ * (2 moles H₂O / 1 mole O₂) = 6.26 moles H₂O

5. Mass of H₂O:

   • Mass of H₂O = 6.26 moles * 18.02 g/mol ≈ 112.8 g

Answer: Approximately 112.8 grams of water are produced.

Stoichiometry Worksheet 1: Mass-Mass Problems (Sample Questions & Answers)

Here's a sample stoichiometry worksheet with detailed answers to help you practice:

(Note: Molar masses are rounded for simplicity. Use more precise values for accurate calculations in a real-world scenario.)

Problem 1: The reaction of aluminum (Al) with hydrochloric acid (HCl) produces aluminum chloride (AlCl₃) and hydrogen gas (H₂):

2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

If 27 grams of aluminum react completely, what mass of hydrogen gas is produced?

Answer:

1. Moles of Al: 27 g / 27 g/mol = 1 mole
2. Mole ratio: 2 moles Al : 3 moles H₂
3. Moles of H₂: 1 mole Al * (3 moles H₂ / 2 moles Al) = 1.5 moles H₂
4. Mass of H₂: 1.5 moles * 2 g/mol = 3 g

Therefore, 3 grams of hydrogen gas are produced.

Problem 2: Consider the reaction of iron (Fe) with oxygen (O₂) to form iron(III) oxide (Fe₂O₃):

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

If 56 grams of iron react completely, what mass of iron(III) oxide is produced?

Answer:

1. Moles of Fe: 56 g / 56 g/mol = 1 mole
2. Mole ratio: 4 moles Fe : 2 moles Fe₂O₃
3. Moles of Fe₂O₃: 1 mole Fe * (2 moles Fe₂O₃ / 4 moles Fe) = 0.5 moles Fe₂O₃
4. Mass of Fe₂O₃: 0.5 moles * 160 g/mol = 80 g

Therefore, 80 grams of iron(III) oxide are produced.

Problem 3: In the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

What mass of ammonia (NH₃) is produced from 14 grams of nitrogen gas (N₂) reacting with excess hydrogen gas?

Answer:

1. Moles of N₂: 14 g / 28 g/mol = 0.5 moles
2. Mole ratio: 1 mole N₂ : 2 moles NH₃
3. Moles of NH₃: 0.5 moles N₂ * (2 moles NH₃ / 1 mole N₂) = 1 mole NH₃
4. Mass of NH₃: 1 mole * 17 g/mol = 17 g

Therefore, 17 grams of ammonia are produced.

Problem 4 (Limiting Reactant): 10 grams of calcium carbonate (CaCO₃) react with 10 grams of hydrochloric acid (HCl) according to the equation:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

What mass of carbon dioxide (CO₂) is produced?

Answer:

1. Moles of Reactants:

   • Moles CaCO₃: 10 g / 100 g/mol = 0.1 moles
   • Moles HCl: 10 g / 36.5 g/mol ≈ 0.27 moles

2. Limiting Reactant: The mole ratio of CaCO₃ to HCl is 1:2. 0.1 moles CaCO₃ require 0.2 moles HCl. Since we have 0.27 moles HCl, HCl is in excess, and CaCO₃ is the limiting reactant.

3. Moles of CO₂: Using the mole ratio from the balanced equation (1:1 for CaCO₃ to CO₂):

   • Moles CO₂ = 0.1 moles

4. Mass of CO₂:

   • Mass CO₂ = 0.1 moles * 44 g/mol = 4.4 g

Therefore, 4.4 grams of carbon dioxide are produced.

Conclusion

Mass-mass stoichiometry problems, while initially challenging, become manageable with a systematic approach. By understanding the steps involved – balancing the equation, converting to moles, using mole ratios, and converting back to grams – you can confidently tackle these problems. Remember to always identify the limiting reactant when multiple reactants are involved. Practice is key to mastering stoichiometry; work through numerous problems, and you'll soon find yourself proficient in this essential area of chemistry. This comprehensive guide, along with the provided worksheet and solutions, provides a strong foundation for further exploration of stoichiometric calculations. Remember to always double-check your work and use precise molar masses for the most accurate results.