Transcript Chapter 12
Stoichiometry CHAPTER 12 The Arithmetic of Equations SECTION 1 SECTION 1 LEARNING TARGETS 12.1.1 – I can explain how balanced equations apply to both chemistry and everyday life. 12.1.2 – I can interpret balanced chemical equations in terms of moles, representative particles, mass, and gas volume at STP. 12.1.3 – I can identify the quantities that are always conserved in chemical reactions. USING EVERYDAY EQUATIONS A balanced chemical equation provides the same kind of quantitative information that a recipe does. USING BALANCED CHEMICAL EQUATIONS Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. EXAMPLE: Tiny tike has decided to make 288 tricycles each day. How many tricycle seats, wheels, and pedals are needed? Stoichiometry – the calculation of quantities in chemical reactions. INTERPRETING CHEMICAL EQUATIONS A balanced chemical equation can be interpreted in terms of different quantities, including; numbers of atoms, molecules, or moles, mass, and volume. NUMBER OF ATOMS A balanced equation indicates that the number and type of each atom that makes up each reactant also makes up each product. NUMBER OF MOLECULES Here the coefficients tell you how many molecules will react and form. Much like how many atoms. MOLES A balanced equation also tells you the number of moles of reactants and products. The coefficients tell you this. You’ll use this most often. MASS A balanced equation obeys the law of conservation of mass. Using a mole relationship you can relate number of moles to mass of either reactants or products. VOLUME If you are at STP the equation tells you about volumes of gases. You can use mole relationships for this also. MASS CONSERVATION IN CHEMICAL REACTIONS Mass and atoms are conserved in every chemical reaction. EXAMPLE: Balance the following equation. ___C2H4(g) + ___O2(g) → ___CO2(g) + ___H2O(g) Interpret the balanced equation in terms of relative numbers of moles, volumes of gas at STP, and masses of reactants and products. Chemical Calculations SECTION 2 SECTION 2 LEARNING TARGETS 12.2.1 – I can construct mole ratios from balanced chemical equations and apply these ratios in stoichiometric calculations. 12.2.2 – I can calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP. WRITING AND USING MOLE RATIOS Mole ratio – a conversion factor derived from the coefficients of a balanced chemical reaction interpreted in terms of moles. In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of products, or between moles reactants. EXAMPLE: 4Al(s) + 3O2(g) → 2Al2O3(s) Write six mole ratios that can be derived from this equation. MOLE-MOLE CALCULATIONS The easiest way to see these is to do an example. W is the unknown, G is the given quantity. a and b are the coefficients from the balanced chemical equation. 4Al(s) + 3O2(g) → 2Al2O3(s) How many moles of aluminum are needed to form 3.7 moles of aluminum oxide? MASS-MASS CALCULATIONS In a reaction things are not measured in moles (no scale does this). Instead things are measured in mass then transferred to moles. STEPS IN SOLVING A MASS-MASS PROBLEM 1. Change the mass of G to moles of G (mass G → mol G) by using the molar mass of G. 2. Change the moles of G to moles of W (mol G →mol W) by using the mole ratio from the balanced equation. 3. Change the moles of W to grams of W (mol W → mass W) by using the molar mass of W. EXAMPLE: Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(aq) How many grams of acetylene are produced by adding water to 5.00g of calcium carbide? OTHER STOICHIOMETRIC CALCULATIONS In a typical stoichiometric problem, the given quantity is first converted to moles. Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the moles are converted to any other unit of measure related to the unit mole, as the problem requires. EXAMPLE: How many molecules of oxygen are produced by the decomposition of 6.54g of potassium chlorate (KClO3)? 2KClO3(s) → 2KCl(s) + 3O2(g) EXAMPLE: The equation for the combustion of carbon monoxide is: 2CO(g) + O2 → 2CO2(g) How many liters of oxygen are required to burn 3.86L of carbon monoxide? Limiting Reagent and Percent Yield SECTION 3 SECTION 3 LEARNING TARGETS 12.3.1 – I can identify the limiting reagent in a reaction. 13.3.2 – I can calculate theoretical yield, actual yield, or percent yield given appropriate information. LIMITING AND EXCESS REAGENTS When cooking you know you need the right amounts of ingredients for the recipe to turn out. In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. Limiting reagent – determines the amount of product that can be formed in the reaction. This is the one used up first in a reaction. The reaction can only “go” until this reactant is completely used up. Excess reagent – reactant that is not completely used up in a reaction. How would the amount of products formed if you started with four molecules of N2 and three molecules H2? EXAMPLE: The equation for the complete combustion of ethene (C2H4) is: C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) If 2.70mol of C2H4 is reacted with 6.30mol of O2, identify the limiting reagent. The reactant that is present in the smaller amount by mass or volume is not necessarily the limiting reagent. EXAMPLE: The heat from an acetylene torch is produced by burning acetylene (C2H2) in oxygen: 2C2H2 + 5O2 → 4CO2 + 2H2O How many grams of water can be produced by the reaction of 2.4 mol C2H2 with 7.4 mol O2? PERCENT YIELD Your grades are usually expressed as a percent right 100 % total Chemists use similar calculations when products are formed based on balanced equations. In theory all reactions would produce at 100%. In reality they don’t. Theoretical yield – the maximum amount of product that could be formed from given amounts of reactants. Actual yield – the amount of product that actually forms when the reaction is carried out. Percent yield – ratio of the actual yield to the theoretical yield expressed as a percent. The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. Percent yield can be lowered by: Impure reactants. Loss of product in filtration or transferring. If reactants or products have not been carefully measured. EXAMPLE: When 84.8g of iron (III) oxide reacts with an excess of carbon monoxide, iron is produced. Fe2O3(s) + 3CO → 2Fe(s) + 3CO2(g) What is the theoretical yield of iron? EXAMPLE: If 50.0g of silicon dioxide is heated with an excess of carbon, 27.9g of silicon carbide is produced. SiO2(s) + 3C(s) → SiC(s) + 2CO(g) What is the percent yield of this reaction?