Curriculum
Y1 · IV · #35Stoichiometry – chemical calculations

Chemical calculations from chemical equations (II)

Introduction

In the last lesson you learned how to use a balanced equation to convert between moles, masses and volumes of reactants and products. Today we go a step further: what happens when you don't have exactly the right amounts of each reactant? This is where the ideas of limiting reactant, theoretical yield and percentage yield come in — essential tools for real laboratory and industrial chemistry (Talbot, p.418).

Why equations don't always react "perfectly"

A balanced equation tells you the ratio in which substances react, not how much of each you actually have in the flask. In most real reactions, one reactant runs out before the other. The reactant that is used up first is called the limiting reactant, because it limits how much product can form. The other reactant, left over at the end, is the excess reactant.

Method for finding the limiting reactant:

  1. Convert all given masses/volumes into moles (n = m / M or n = cV).
  2. Divide each mole amount by its coefficient in the balanced equation.
  3. The reactant with the smallest value is the limiting reactant.

Worked Example 1

Reaction: N₂(g) + 3H₂(g) → 2NH₃(g)

Suppose you have 2.0 mol N₂ and 3.0 mol H₂.

  • N₂: 2.0 / 1 = 2.0
  • H₂: 3.0 / 3 = 1.0

H₂ gives the smaller number, so H₂ is the limiting reactant. All further calculations (mass of NH₃ formed, leftover N₂, etc.) must be based on H₂, not N₂.

Theoretical yield vs

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Source excerpts

From Chemistry for the IB Diploma 3e · Talbot

p.418relevance 16.7

R2: How much, how fast and how far? 406 How much? The amount of chemical changeR2.1 • How are chemical equations used to calculate reacting ratios? Guiding question SYLLABUS CONTENT By the end of this chapter, you should understand that:  chemical equations show the ratio of reactants and products in a reaction  the mole ratio of an equation can be used to determine:  the ma…

p.525relevance 15.3

HL ONLY R2.3 How far? The extent of chemical change 513 Going further Calculations using the quadratic formula For the esterification reaction: CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l) calculate the amount of ethyl ethanoate that formed at equilibrium when 1.0 mole of ethanol reacted with 2.0 moles of ethanoic acid at 373 K. T he value of K is 4.0 at this temperature. Le…

p.525relevance 14.1

with 1.0 mole of ethanol. Therefore the amount of ethyl ethanoate at equilibrium is 0.85 moles. The IB syllabus states that calculations of this type (involving the solution of quadratic equations) will not be set. However, questions 44 and 45 lead to equations that include perfect squares. It is worth remembering that these can be solved by taking the square root of each side …

p.82relevance 12.6

re actions. Common mistake Misconception: Mass is not conserved during a chemical reaction. The products of chemical reactions need not have the same mass as the re actants. Mass is always conserved during a chemical reaction. The products of chemical reactions must have the same mass as the reactants since atoms cannot be created or destroyed during a chemical reaction. Chemic…

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP
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