C5.1 Monitoring Chemical Reactions

Maximum mass it is possible to make from a given mass of reactants Mass limiting reactant (g) ———————————— x sum of Mr for products Limiting reactant Mr

Actual yield ———————— x 100 Theoretical yield Answer is a percentage Cannot be more than 100%

Varies from 0% to 100%. Less than 100% because: Reactants can react in a different way than expected Reaction may not go to completion Some of the reactants present do not react Often happens in reversible reactions You may lose some of the product when you separate it from the reaction mixture and purify it, such as filtration

Sum of Mr of desired products ——————————————— x 100 Sum of Mr of all products Answer as a percentage Cannot be more than 100% Percentage of reactants turned into useful products

Manufacturers in the chemical industry must profit from producing substances or they go out of business Chemical engineers must consider many factors choosing a reaction pathway: Yield of product Atom economy of the reaction Usefulness of by products Rate of reaction Equilibrium position if is a reversible reaction

Byproduct = substance formed in a reaction in addition to desired product Can sometimes be useful and sold improving atom economy

1

1

Multiply by 1000

. Mass of solute (g) Concentration in g/dm3 = ————————————— Volume of solution (dm³) Amount of solute (mol) Concentration in mol/dm3 = ———————————— Volume of solution (dm³)

Add a KNOWN volume of alkali in a conical flask and few drops of SINGLE indicator like phenolphthalein Then add acid to flask from a burette bit by bit At end point when the alkali is neutralised and the indicator first changes colour stop adding acid

Solution with known accurate concentration

Use measuring cylinder to measure 25 cm³ of alkali solution for the conical flask, but a volumetric pipette is more accurate (Need a pipette filler to use a volumetric pipette safely) Burette must be clamped vertically so you obtain accurate readings When you have filled the burette with acid, and added a few drops of indicator to the alkali in the conical flask, you are ready Eye protection should be worn

Take an initial burette reading and a final burette reading at the end-point Difference between these two readings is the titre is volume of acid added to the alkali in the flask Record your readings to two decimal places, ending in 0 if the bottom of the meniscus is on a burette line, or 5 if the meniscus is between two lines To obtain a repeatable tire you need to use following techniques: Swirl the flask during a titration to mix its contents First titration is usually a rough run, done quickly so that you get an idea of what the titre is In later runs can quickly add the acid to within a few cm³ of the rough titre, then add the acid drop by drop Repeat titration until you obtain at least two concordant titres: titres that are within 0.1 cm³ of each other allows you to calculate a mean tire that has high precision Eye protection should be worn

When you have carried out a titration, you will know: The two reactants used (an acid and an alkali) the volume and concentration of one of the reactants the volume, but not the concentration, of the other reactant With this information, and the balanced equation, you can calculate the unknown concentration

Example: In a titration, 25.0 cm3 of 0.040 mol/dm3 barium hydroxide solution reacted with 21.6 cm3 of hydrochloric acid. Find the concentration of the hydrochloric acid in mol/dm3. Balanced equation: Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H2O(l) Create table of information: (information already known from the question in 𝕥𝕙𝕚𝕤 font) then fill it in using information. Ba(OH)₂ 2HCl Volume(cm³) 𝟚𝟝.𝟘 𝟚𝟙.𝟞 Volume(dm³) 25.0 / 1000 = 0.025 21.6 / 1000 = 0.0216 Concentration (mol/dm³) 𝟘.𝟘𝟜𝟘 Moles = concentration x volume 0.002 = concentration x 0.0216 0̳.̳0̳9̳̳̇2̳̳̇5̳̳̇ ̳m̳o̳l̳ ̳/̳ ̳d̳m̳³̳ ̳=̳ ̳c̳o̳n̳c̳e̳n̳t̳r̳a̳t̳i̳o̳n̳ ̳ Moles (in titration N = 0.04 x 0.025 1 : 2 ratio of Ba(OH)₂ : 2HCl so, questions use N = 0.001 mol 0.002 mol formula: N = conc x vol

1 mole of any substance in gas state occupies the same volume at the same temperature and pressure At room temperature and pressure: 24 dm³/mol (24 000 cm³/mol)

Moles of gas = Volume of gas (dm³) / 24 dm³ Moles of gas = Volume of gas (cm³) / 24 cm³ Moles of gas = volume / molar volume

Step 1: Calculate the amount of limiting product: Ar = 65.4 Moles of zinc = mass (g) / Ar = 1.31 g / 65.4 g/mol = 0.02 mol Step 2: Use balanced equation to calculate the amount of hydrogen produced 1 mol Zn produces 1 mol H2 so 0.02 mol of Zn produces 0.0200 mol of H2 Step 3: Use your answer to Step 2, and Vm to calculate the volume of hydrogen: volume in dm3 = amount in mol × 24 dm³/mol = 0.0200 mol × 24 dm³/mol = 0.48 dm³

1 Fill the measuring cylinder with water and while keeping its mouth underwater turn upside down 2 Clamp the measuring cylinder securely 3 Mix the reactants in a conical flask and attach a delivery tube 4 Measure the volume of gas produced in the reaction by recording the end reading on the measuring cylinder In this investigation, you need to keep the mouth of the measuring cylinder underwater to stop air getting in or water getting out Either of these would affect the volume reading on the measuring cylinder Eye protection should be worn throughout this practical Or just use a gas syringe

Volume of acid added to an alkali, to fully neutralise it

Mol/ dm³ x Mr —> g/dm³