Year 11 Chemistry: Chapter 15:~ Calculations in Chemistry: Stoichiometry Notes from ‘Chapter 4: Relative atomic mass and mole’ will help Reacting Quantities A chemist working for a drug company needs to be precise when weighing the reactants required to prepare a particular compound. Chemists involved in the preparation and analysis of a wide range of compounds rely heavily on calculations of masses of reactants and products. Calculations such as these are based on balanced, full chemical equations for each reaction. All particles that existed at the start of a chemical reaction must be accounted for after the reaction is complete. 2H2(g) + O2(g) 2H2O(g) This equation gives the following information: It identifies the reactants and products. It identifies the states of the reactants and products Because it is balanced, it gives the ratio in which the substances react. The coefficients in a balanced equation therefore give us the ratio of the amounts (in mol) of reactants and products. This mole ratio will be used in calculating masses of reactants and products. 15.1 Calculations based on amount of a reactant or product Mass-Mass Stoichiometry In these problems, the mass of one of the reactants or products is known and is used to determine the mass of one of the other reactants or products involved in the reaction. Step 1: Write a balanced equation for the reaction. Step 2: Calculate the amount (in mol) of the substance with the known mass. m n = M Step 3: Use the mole ratio from the equation to calculate the amount (in mol) of the required substance. Step 4: Calculate the mass required m = nM microgram milligram gram kilogram tonne g mg g kg t Example: Iron rusts slowly when exposed to air and water. The reaction involved in rusting can be represented by the equation below. What mass of iron oxide is produced when 120g of iron completely in air? Step 1: Write a balanced equation 4Fe(s) + 3O2(g) 2Fe2O3(s) Step 2: Calculate the amount (in mol) of the substance with the known mass. m n = M M(Fe2O3) = Step 3: Mole ratio n (unknown) n (know) = Step 4: Calculate the mass required n = m x M Example: A phosphorus manufacturer is to extract 1.00 tonne of phosphorus per day by the process given by the equation. 2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) P4(s) + 10CO(g) + 6CaSiO3(s) Calculate the mass required daily of: a) calcium phosphate b) silicon dioxide a) Calcium phosphate Step 1: Write the balanced equation Step 2: Calculate mole of known substance Step 3: mole ratio Step 4: Calculate the mass required b) Silicon Dioxide Step 1: Write the balanced equation Step 2: Calculate mole of known substance Step 3: mole ratio Step 4: Calculate the mass required Stoichiometry involving solutions Many chemical reactions occur between reactants in solution. The amount of a reactant present in a solution can be determined from the volume of the solution used and its concentration, by using the relationship developed in Chapter 11. n(mol) = c (mol L-1) x V (L) Example: Precipitation reactions occur when two solutions containing soluble reactants are mixed together to produce an insoluble product. This product is collected as a precipitate. What mass of barium sulfate is produce when 100 mL of a 1.52M solutions of barium nitrate reacts completely with a sodium sulfate solution. Step 1: Write a balanced equation Step 2: Calculate mole of known solution Step 3: mole ratio Step 4: Calculate mass of unknown solution Example: What volume of 0.100M sulfuric acid reacts completely with 17.8 mL of 0.152M potassium hydroxide. Step 1: Write a balanced equation Step 2: Calculate mole of known solution Step 3: mole ratio Step 4: Calculate volume of unknown solution Questions: Week 6 Year 11 Chemistry: Chapter 15:~ Calculations in Chemistry: Stoichiometry 15.2 Calculations based on amount of two reactants In the previous examples the quantity of one substance was given and used to calculate the quantity of other reactants or products. The previous calculations assumed that any other reactants were present in sufficient quantities to react completely with all of the given substance. In the following problems, fixed quantities of two different reactants are mixed. It is likely that one of the reactants will be used up before the other. The reaction stops when one reactant is used up (limiting reagent), even though some of the other substance is unreacted. The other reactant is said to be in excess (excess reagent). 2H2(g) + O2(g) 2H2O(g) Container A: holds 8 hydrogen molecules and 6 oxygen molecules, they react Container B: 8 water molecules and two oxygen molecules left. 2H2(g) + O2(g) 2H2O(g) Initial amount 8 mol React’n change according to eq’n 8 mol Final amounts 0 mol After the reaction: no hydrogen gas remains, since this is the limiting reagent. The amount of water vapour formed is limited by the amount of hydrogen gas available. 2 mol of oxygen gas remains unreacted. Oxygen gas is in excess in this example. Example: A gaseous mixture of 25.0 g of hydrogen gas and 100.0 g of oxygen gas are mixed and ignited. The water produced is collected and weighed. What is the expected mass of water produced? Step 1: Write a balanced chemical equation Step 2: Calculate the amount of each reactant. M(H2) = m m M(O2) = n (H2) = M n(O2) = M Step 3: Identify the reagent in excess, using mole ratio. Use the reactant with the smallest mole. Step 4: Calculate the amount of the limiting reagent to determine the amount (in mol) of the required product. Step 5: Calculate mass of product Example: 2.50 g of aluminium is mixed with 5.00 g of iodine and allowed to react. a) What mass of aluminium iodide would be produced? b) What is the mass of the reactant in excess? Step 1: Write a balanced chemical equation Step 2: Calculate the amount of each reactant Step 3: Identify reactant in excess. Step 4: use the limiting reagent to determine the quantity required to produce. Step 5: Calculate the mass. a) aluminium iodide b) mass of aluminium in excess Questions: 9, 10, 11, 34, 35, 36 & 39 Volumetric Analysis - Titrations There are many situations when it is essential to know the exact amount of acid or base in a substance. The concentration of solutions of acids and bases can be determined accurately by a technique called volumetric analysis. This involves reacting the solution of unknown concentration with a solution of accurately known concentration (standard solution). Volumetric flask: Pipette: Aliquot: Burette: Titre: Equivalence point: is where the acid and base are neutralised. End point: indicator changes colour close to the equivalence point. Since both acid and base solutions are often colourless, there needs to be some way to determine when the reaction is complete. This is done by adding an indicator that changes colour at (or very close) to the equivalence point of the titration (end point). Rinsing equipment: Experiment 54 as example Pipette: water then, sample from volumetric flask Burette: water then, solution that is being placed in the burette Conical flask: distilled water Volumetric flask: distilled water. Questions: 12, 13, 40, 41 & 43. Questions: 1, 2, 3, 4, 5, 6, 7, 8 (should already be completed), 15, 16, 17, 19, 21, 22, 24, 26, 28, 29, 30, 31, 33, & 45.
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