Eduqas A-Level Chemistry A410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Specified Practical Work
Practical Skills and Techniques
Mathematical Skills in Chemistry
Chemical Calculations
Stoichiometry and reacting quantities
- 3 things to know
- Components 1 and 2
Overview
Stoichiometry of a reaction, steps of a reacting-mass calculation and term limiting reagent.
What you need to know
Answer each one before you reveal it.
What is meant by the stoichiometry of a reaction?
What mass of calcium oxide is formed when 25.0 g of calcium carbonate is fully decomposed?
Questions from this subtopic
Each one with the model answer underneath.
- 1
What is meant by the stoichiometry of a reaction?
Model answer
Stoichiometry is the fixed whole-number ratio in which the substances react and are formed, given by the coefficients of the balanced equation.
- 2
List the steps of a reacting-mass calculation.
Model answer
Write the balanced equation, convert the known mass into moles, apply the mole ratio from the equation to find the amount of the required substance, and convert that amount back into a mass.
- 3
What mass of calcium oxide is formed when 25.0 g of calcium carbonate is fully decomposed?
Model answer
The amount of CaCO₃ is 25.0/100.1 = 0.250 mol, and since the ratio is one to one the mass of CaO is 0.250 × 56.1 = 14.0 g.
- 4
What volume of carbon dioxide, measured at room temperature and pressure, is released when 0.250 mol of calcium carbonate decomposes?
Model answer
One mole of CO₂ forms per mole of carbonate, so the volume is 0.250 × 24.5 = 6.13 dm³.
- 5
Define the term limiting reagent.
Model answer
The limiting reagent is the reactant that is completely used up first and therefore fixes the maximum amount of product that can form.
- 6
How is the limiting reagent identified when the amounts of both reactants are known?
Model answer
Divide the amount in moles of each reactant by its coefficient in the balanced equation, and the reactant giving the smallest result is the limiting one.
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