Ch 4 · Reacting masses and volumes of solutions and gases · AS & A Level Chemistry 9701

Topic 4 of 4 in Atoms, molecules and stoichiometry — AS & A Level Chemistry 9701 — 56 lessons.

Introduction

A balanced equation is a ratio of moles, and every quantitative problem in chemistry — from reacting masses to gas volumes to titrations — is solved by converting the given data into moles, applying that ratio, and converting back to the quantity required. This lesson builds that method systematically, including limiting reagents, percentage yield and appropriate significant figures.

Key ideas — Reacting masses and percentage yield

To find a reacting mass: convert the given mass to moles using n = m/M, use the mole ratio from the balanced equation to find moles of the required substance, then convert back to mass using m = n × M. The percentage yield compares the actual mass of product obtained with the maximum theoretical mass calculated from stoichiometry:

🧑‍🏫 Teacher's explanation

Think of stoichiometry like a recipe for baking cookies. If a recipe says you need 2 eggs to make 12 cookies, and you have 6 eggs, you can theoretically make 36 cookies. However, in a real kitchen, you might drop some dough, burn a few, or accidentally use slightly too much flour, meaning you actually only end up with 30 cookies. The 'theoretical yield' is that perfect 36, and the 'actual yield' is the 30 you actually hold in your hand.

📖 Definitions

🔍 Worked example — Easy

Calculate the theoretical yield of MgO when 24.0 g of Mg reacts completely with excess O_2. (M_r of Mg = 24.0, O = 16.0)

🔍 Worked example — Standard

A reaction produces 15.0 g of CO_2. The theoretical yield calculated was 20.0 g. Calculate the percentage yield.

🔍 Worked example — Exam-style

Calculate the mass of Fe_2O_3 produced when 11.2 g of Fe reacts with excess O_2. The actual yield obtained is 14.0 g. Calculate the percentage yield. (M_r of Fe = 56.0, O = 16.0)

⚠️ Common errors

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🎯 Exam question

Key ideas — Volumes of gases

At room temperature and pressure (r.t.p.), one mole of any gas occupies a molar volume of 24.0\ dm^3 (24 000 cm³), so

🧑‍🏫 Teacher's explanation

Think of gas molecules like tiny, energetic bouncy balls. Even though they are incredibly small, they take up a huge amount of space because they are constantly flying around and pushing against each other. Because they are so spread out, the actual volume they occupy depends heavily on how many 'balls' (moles) you have and how much room they are allowed to take up.

📖 Definitions

🔍 Worked example — Easy

Calculate the volume of 0.50 moles of CO_2 gas at r.t.p. in cm^3.

🔍 Worked example — Standard

A reaction produces 1.2 dm^3 of H_2 gas at r.t.p. How many moles of H_2 were produced?

🔍 Worked example — Exam-style

Magnesium reacts with hydrochloric acid according to the equation: Mg(s) + 2HCl(aq) arrow MgCl_2(aq) + H_2(g). If 0.24 g of Mg reacts completely, calculate the volume of H_2 gas produced at r.t.p. in dm^3. (Ar: Mg = 24)

⚠️ Common errors

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🎯 Exam question

Key ideas — Volumes and concentrations of solutions

The concentration of a solution, c, is related to the amount dissolved and the volume of solution by

🧑‍🏫 Teacher's explanation

Think of concentration as the 'crowdedness' of a room. If you have 10 people in a tiny bathroom, the concentration of people is very high. If those same 10 people are in a football stadium, the concentration is very low. In chemistry, we aren't counting people; we are counting particles (moles) inside a specific volume of liquid.

📖 Definitions

🔍 Worked example — Easy

Calculate the concentration in mol dm^-3 of a solution containing 0.5 moles of NaOH in 250 cm^3 of solution.

🔍 Worked example — Standard

A student titrates 25.0 cm^3 of 0.100 mol dm^-3 HCl with NaOH. The NaOH requires 20.0 cm^3 to reach the endpoint. Calculate the concentration of the NaOH. The reaction is: HCl + NaOH arrow NaCl + H_2O.

🔍 Worked example — Exam-style

Calculate the volume, in cm^3, of 0.50 mol dm^-3 H_2SO_4 required to react completely with 25.0 cm^3 of 0.20 mol dm^-3 KOH. The reaction is: H_2SO_4 + 2KOH arrow K_2SO_4 + 2H_2O.

⚠️ Common errors

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🎯 Exam question

Key ideas — Limiting reagent and reagent in excess

When two reactants are mixed, the limiting reagent is the one that is completely used up first, and it determines the maximum amount of product that can form; the reagent in excess is left over, partly unreacted, once the limiting reagent is exhausted. To identify the limiting reagent: calculate the moles of each reactant available, divide each by its coefficient in the balanced equation, and the reactant giving the smaller value is limiting. The amount of excess reagent remaining is found by calculating how much of it reacts with all of the limiting reagent (via the mole ratio), and subtracting this from the amount originally present.

🧑‍🏫 Teacher's explanation

Imagine you are making sandwiches. Each sandwich requires 2 slices of bread and 1 slice of cheese. If you have 10 slices of bread and 2 slices of cheese, you can only make 2 sandwiches. Even though you have plenty of bread left, you ran out of cheese. In this scenario, cheese is the 'limiting reagent' because it dictates how many sandwiches you can make, and bread is the 'excess reagent' because you have leftovers.

📖 Definitions

🔍 Worked example — Easy

Magnesium reacts with hydrochloric acid according to the equation: Mg + 2HCl arrow MgCl_2 + H_2. If 0.50 g of Mg is reacted with 0.40 g of HCl, identify the limiting reagent. (A_r: Mg=24, Cl=35.5, H=1)

🔍 Worked example — Standard

Calculate the mass of excess Mg remaining when 0.50 g of Mg reacts with 0.40 g of HCl.

⚠️ Common errors

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🎯 Exam question

Key ideas — Significant figures and stoichiometric ratios

A calculated answer should be reported to the same number of significant figures as the least precise piece of data given in the question, since a calculation cannot be more precise than its least precise input; reporting more figures overstates the precision of the result, and reporting fewer discards information unnecessarily. When experimental reacting-quantity data (such as titration results) is used to deduce an unknown mole ratio, the moles of each reactant are calculated independently and then compared as a ratio, which is rounded to the nearest simple whole-number ratio to give the stoichiometric coefficients in the equation.

🧑‍🏫 Teacher's explanation

Think of stoichiometry as a recipe in a kitchen. If a recipe for 1 batch of cookies requires 2 eggs and 500g of flour, and you have 4 eggs and 1000g of flour, you can make exactly 2 batches. The 'ratio' is the relationship between the ingredients. In chemistry, the balanced equation tells us this ratio for atoms and molecules.

📖 Definitions

🔍 Worked example — Easy

Calculate the mass of H_2 produced when 0.50 mol of Mg reacts completely with excess HCl. The equation is Mg + 2HCl arrow MgCl_2 + H_2. (M_r of H_2 = 2)

🔍 Worked example — Standard

A reaction involves 2H_2 + O_2 arrow 2H_2O. If 10.0 g of H_2 reacts with 10.0 g of O_2, identify the limiting reagent and calculate the mass of H_2O produced. (M_r of H_2=2, O_2=32, H_2O=18)

🔍 Worked example — Exam-style

In a titration, 25.00 cm^3 of NaOH (aq) reacts with 20.0 cm^3 of H_2SO_4 (aq). The volume of NaOH required to neutralize 20.0 cm^3 of H_2SO_4 is 25.00 cm^3. Calculate the concentration of H_2SO_4 in mol dm^-3. The equation is 2NaOH + H_2SO_4 arrow Na_2SO_4 + 2H_2O.

⚠️ Common errors

🧠 Quick check 1

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🎯 Exam question

Worked example

Question. 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution is titrated with 0.0500 mol dm⁻³ sulfuric acid, H_2SO_4, according to the equation 2NaOH(aq) + H_2SO_4(aq) arrow Na_2SO_4(aq) + 2H_2O(l) (a) Calculate the moles of NaOH used. [1] (b) Calculate the volume of sulfuric acid required to exactly neutralise the sodium hydroxide. [3]

Ch 4 · Reacting masses and volumes of solutions and gases · AS & A Level Chemistry 9701

All chapters in Atoms, molecules and stoichiometry