Reaction Stoichiometry

Reaction Stoichiometry

Learning Objectives

  • Explain the concept of stoichiometry as it pertains to chemical reactions.
  • Use balanced chemical equations to derive stoichiometric factors.
  • Perform calculations involving mass, moles, and solution molarity.

A balanced chemical equation provides a quantitative assessment of the relationships between the amounts of substances consumed and produced by a reaction. These quantitative relationships are known as the reaction’s stoichiometry. In simple terms, stoichiometry is like a recipe in cooking—it tells you how much of each ingredient you need and how much product you'll get. It's based on the law of conservation of mass, ensuring that atoms are neither created nor destroyed in a reaction.

1. Stoichiometric Factors

The coefficients in a balanced equation allow us to derive stoichiometric factors—ratios that permit computation of the desired quantity of one substance relative to another. These factors are essentially conversion tools, like miles to kilometers, but for chemicals in a reaction.

Allied Health Insight: Synthesis Ratios
In pharmaceutical manufacturing, stoichiometric factors ensure that the correct ratio of precursors is used to maximize the yield of an active drug ingredient while minimizing toxic byproducts.

Consider the production of ammonia:

N₂(g) + 3H₂(g) → 2NH₃(g)

From this, we can derive factors such as:

  • 2 mol NH₃ / 3 mol H₂ (To find product from reactant)
  • 3 mol H₂ / 1 mol N₂ (To find one reactant from another)
  • 1 mol N₂ / 2 mol NH₃ (To find reactant from product)

More details: The coefficients (1 for N₂, 3 for H₂, 2 for NH₃) represent the mole ratios. So, for every 1 mole of N₂, you need 3 moles of H₂ to produce 2 moles of NH₃. These ratios are fixed and come directly from the balanced equation.

Step-by-Step Example: Mole-to-Mole Conversion (Ammonia Synthesis)

Problem: How many moles of NH₃ can be produced from 4.5 moles of H₂?

  1. Write the balanced equation: N₂(g) + 3H₂(g) → 2NH₃(g)
  2. Identify the stoichiometric factor: From the equation, 3 mol H₂ produces 2 mol NH₃, so the ratio is 2 mol NH₃ / 3 mol H₂.
  3. Set up the calculation: 4.5 mol H₂ × (2 mol NH₃ / 3 mol H₂)
  4. Compute: (4.5 × 2) / 3 = 9 / 3 = 3 mol NH₃

Answer: 3 moles of NH₃.

2. Moles of Reactant Required

Example: Phosphoric Acid and Dental/Bone Health
Calcium hydroxide and phosphoric acid react to produce calcium phosphate, a compound related to bone and tooth mineral: 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O

Problem: How many moles of Ca(OH)₂ are required to react with 1.36 mol of H₃PO₄?

Calculation:
1.36 mol H₃PO₄ × (3 mol Ca(OH)₂ / 2 mol H₃PO₄) = 2.04 mol Ca(OH)₂

More details: This is a direct mole-to-mole calculation using the ratio from the balanced equation. The factor 3/2 comes from the coefficients: 3 for Ca(OH)₂ and 2 for H₃PO₄.

Step-by-Step Example: Another Mole Requirement (Water Formation)

Balanced equation for water: 2H₂(g) + O₂(g) → 2H₂O(l)

Problem: How many moles of O₂ are needed to react with 5 moles of H₂?

  1. Write the balanced equation: 2H₂ + O₂ → 2H₂O
  2. Find the ratio: 1 mol O₂ / 2 mol H₂
  3. Calculate: 5 mol H₂ × (1 mol O₂ / 2 mol H₂) = 2.5 mol O₂

Answer: 2.5 moles of O₂.

3. Mass-to-Mass Relationships

In clinical practice, we measure mass (grams) rather than moles. Practical stoichiometry requires converting mass to moles, using the stoichiometric factor, and then converting back to mass. This involves using molar masses (from the periodic table) to convert between grams and moles.

Example: Antacid Production (Milk of Magnesia)
Milk of magnesia, Mg(OH)₂, is produced by reacting sodium hydroxide with magnesium chloride: MgCl₂(aq) + 2NaOH(aq) → Mg(OH)₂(s) + 2NaCl(aq)

Problem: What mass of NaOH is required to produce 16 g of Mg(OH)₂?

Steps:

  1. Grams to Moles: 16 g Mg(OH)₂ ÷ 58.3 g/mol = 0.274 mol Mg(OH)₂
  2. Mole Ratio: 0.274 mol × (2 mol NaOH / 1 mol Mg(OH)₂) = 0.548 mol NaOH
  3. Moles to Grams: 0.548 mol NaOH × 40.0 g/mol = 22 g NaOH

More details: Molar mass of Mg(OH)₂ is Mg (24.3) + 2O (32) + 2H (2) = 58.3 g/mol. For NaOH: Na (23) + O (16) + H (1) = 40 g/mol. Always round appropriately based on given data.

Step-by-Step Example: Mass-to-Mass (Combustion of Methane)

Balanced equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Problem: What mass of O₂ is needed to burn 8 g of CH₄? (Molar mass CH₄ = 16 g/mol, O₂ = 32 g/mol)

  1. Convert given mass to moles: 8 g CH₄ ÷ 16 g/mol = 0.5 mol CH₄
  2. Use mole ratio: 0.5 mol CH₄ × (2 mol O₂ / 1 mol CH₄) = 1 mol O₂
  3. Convert moles to mass: 1 mol O₂ × 32 g/mol = 32 g O₂

Answer: 32 g of O₂.

4. Calculations Involving Solution Molarity

Molarity (M) is moles of solute per liter of solution. In stoichiometry with solutions, we use volume and molarity to find moles (moles = M × L), then apply stoichiometric factors.

Step-by-Step Example for Beginners: Acid-Base Neutralization

Balanced equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Problem: How many liters of 0.5 M NaOH are needed to neutralize 0.2 L of 1 M HCl?

  1. Find moles of given: Moles HCl = 1 M × 0.2 L = 0.2 mol HCl
  2. Use mole ratio: 0.2 mol HCl × (1 mol NaOH / 1 mol HCl) = 0.2 mol NaOH
  3. Find volume: Volume NaOH = moles / M = 0.2 mol / 0.5 M = 0.4 L

Answer: 0.4 liters of NaOH solution.

Another Example: Precipitation Reaction

Balanced equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Problem: What volume of 0.1 M AgNO₃ is required to react with 50 mL of 0.2 M NaCl? (Convert mL to L: 50 mL = 0.05 L)

  1. Moles NaCl = 0.2 M × 0.05 L = 0.01 mol NaCl
  2. Mole ratio: 0.01 mol NaCl × (1 mol AgNO₃ / 1 mol NaCl) = 0.01 mol AgNO₃
  3. Volume AgNO₃ = 0.01 mol / 0.1 M = 0.1 L (or 100 mL)

Answer: 0.1 liters of AgNO₃ solution.

Chemistry in Everyday Life: Airbags

Automotive airbags deploy via the rapid decomposition of sodium azide (NaN₃): 2NaN₃(s) → 3N₂(g) + 2Na(s)

Stoichiometry is used to calculate the exact mass of NaN₃ (~100 g) required to generate enough nitrogen gas (~50 L) to inflate the bag in 0.03–0.1 seconds. This involves gas stoichiometry (using ideal gas law for volumes), but at basic level, it's about ensuring the right mass produces the right moles of gas.