CHAPTER 10 Chemical Equations Calculations

Transcription

1 CHAPTER 10 Chemical Equations Calculations Suppose you are given the equation below, 2NO (g) + O 2 (g) uv 2NO 2 (g) The above equation is interpreted as 2 molecules of NO (nitrogen monoxide) react with 1 molecule of O 2 (oxygen) to produce 2 molecules of NO 2 (nitrogen dioxide). These coefficients indicate relative number of reactant and product molecules. Interpretation of chemical equation coefficients For the General equation 2A + 3B C + 2D The ratio of molecules is 2 : 3 : 1 : 2 The ratio of moles is 2 : 3 : 1 : 2 The ratio of volumes of gas is 2 : 3 : 1 : 2 The ratio of Avogadro s number is 2 : 3 : 1 : 2 VERIFYING THE CONSERVATION OF MASS LAW Law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. The combined mass of reactants must be equal to the combined mass of products. Example, 2 NO (g) + O 2 (g) 2NO 2 (g) 2(30.01g) + 1(32.00g) 2(46.01g) 92.02g 92.02g MOLE-MOLE RELATIONSHIPS Considering the reaction below: N 2 (g) + O 2 (g) 2NO (g) 1 mol of N2 reacts with 1 mol of O 2, the corresponding mole ratios are: 1 mol N 2 = 1 mol O 2 1 mol O 2 1 mol N 2 1 mol of N 2 produces 2 mol of NO, the corresponding mole ratios are: 1 mol N 2 = 2 mol NO 2 mol NO 1 mol N 2 1 mol of O 2 produces 2 mol of NO, the corresponding mole ratios are: 1 mol O 2 = 2 mol NO 2 mol NO 1 mol O 2 TYPES OF STOICHIOMETRY PROBLEMS. The term stoichiometry is used to refer to the relationship between quantities in a chemical reaction according to the balanced chemical

2 equation. The word is coined from Greek words meaning element measure. Stoichiometry problems can be classified as one of these basic types:massmass, mass- volume, volume- volume problems. MASS- MASS PROBLEMS. In this type of calculation, the unknown mass of substance is calculated from given mass of a reactant or product in a chemical equation as follows: Convert the given mass of substance to moles using the molar mass of the substance as a unit factor. Convert the moles of the given to moles of the unknown using the coefficients in the balanced equation. Convert the moles of the unknown to gram using the molar mass of the substance as a unit factor. Example: calculate the mass of Aluminium, considering the high temperature conversion of 14.4g of iron (II) oxide to elemental iron with aluminium metal. Given equation 3Fe (l) + 2Al (l) 3Fe(l) + Al 2 O 3(l) MM FeO = 71.85g/mol 14.4g FeO x 1 mol FeO = mol FeO 71.85g FeO From the equation, 3 mol of FeO = 2 mol of Al mol of FeO x 2 mol Al = mol Al 3 mol of FeO MM of Al = 28.98g/mol (using the unit factor) 0.134mol Al x 28.98g Al = 3.60g Al 1 mol Al 14.4g of FeO reacts with 3.60g of Al. MASS- VOLUME PROBLEMS. This involves an unknown volume of substance is calculated from given mass of reactant or product in a chemical equation. Convert the given mass of substance to moles using the molar mass of the substance as a unit factor. Convert the moles of the given to moles of the unknown using coefficients in the balanced equation. Convert the moles of the unknown to liters using the molar volume of a gas as a unit factor. At standard temperature and pressure (STP), the molar volume of a gas is 22.4 l/mol.

3 Example, Consider the reaction of 0.165g of aluminium metal with dilute hydrochloric acid, the balanced equation, 2Al (s) + 6HCl (aq) 2AlCl 3 (aq) + 3H 2 (g) Calculate the volume of hydrogen gas produced by the reaction at STP. Number of mol of Al (mm of Al = 26.98g/mol) 0.165g Al x 1 mol Al = mol Al 26.98g Al Number of moles H 2, using the coefficients in balanced equation, mol Al x 3 mol H 2 = mol H 2 2 mol Al Volume of H 2, applying molar volume, 22.4l/1 mol, mol H 2 x 22.4L = 0.250L of H 2 1 mol H 2 VOLUME- VOLUME PROBLEMS In a volume- volume problem, we take advantage of Gay- Lussac s law of combining volumes which states that volumes of gases, under similar conditions, combine in small whole number ratios. H 2 (g) + Cl 2 (g) 2HCl (g) 10mL 10mL 20mL 1 : 1 : 2 2H 2 (g) + O 2 (g) 2H 2 O(g) 100mL 50mL 100mL 2 : 1 : 2 Since the volumes of gases combine in the same whole number ratio as the coefficients in the balanced equation, we can solve this type of problem in one conversion step. We simply multiply the given volume by the ratio of the coefficients in the equation. Example, calculate the liters of oxygen gas that react with 37.5 L of sulfur dioxide, in the following reaction: 2SO 2 (g) + O 2 (g) Pt / 2SO 3 (g) 2L of SO 2 = 1L of O 2 ( from the equation) 37.5L of SO 2 x 1L of O 2 = 18.8L of O 2 2L of SO 2 If we need to calculate the volume of SO3 produced, 2L of SO 2 = 2L of SO 3 ( from the equation) 37.5L of SO 2 x 2L of SO 3 = 37.5L of SO 3 2L of SO 2

4 LIMITING REACTANT CONCEPT A limiting reactant is the reactant that controls the amount of products formed. We need to identify the limiting reactant in a reaction in order to correctly calculate the amount of product. Suppose we have a reaction, Fe + S FeS 0.25mol 3.00mol 0.00mol 0.25mol 0.25mol 0.25mol Since the reaction is ratio 1: 1 : 1, we can conclude that FeS will be 0.25mol. Thereby leaving sulphur with excess of ( = 2.75 mol). We can therefore conclude that Fe is the limiting reagent. Example: A 5.00mol sample of iron(iii) oxide is heated with 5.00mol aluminium metal and converted to molten iron. Identify the limiting reactant, and calculate the number of moles of iron produced, given the equation: Fe 2 O 3(l) + 2Al (l) 2Fe (l) (l) + Al 2 O 3(l) Calculating the number of moles of Fe, 5.00mol Fe 2 O 3 X 2mol Fe = 10 mol Fe 1 mol Fe 2 O mol Al X 2mol Fe = 5 mol Fe 2 mol Al Therefore the limiting reactant is Al and the number of mole of Fe produced was 5 mol. When the mass is given, we determine the limiting reactant as follows: I. Calculating the mass of product that can be produced from the first reactant Calculate the moles of reactant. Calculate the moles of product. Calculate the mass of product. II. Calculate the mass of product that can be produced from second Reactant. Calculate the moles of reactant. Calculate the moles of product calculate the mass of product III. State the limiting reactant and the corresponding mass of product Formed. The limiting agent gives the least amount product. The actual mass of product obtained from the chemical reaction is the lesser of the two product masses calculated in steps I and II. Suppose we wish to find

5 how much molten iron is produced from the reaction of 25.0g of FeO with 25.0g of Al. 3FeO (l) + 2Al (l) 3Fe (l) + Al 2 O 3 (l) g FeO to mol FeO to mol Fe to g Fe Steps: Ia Ib Ic Mm of Fe = 55.85g/mol and Mm of FeO = 71.85g/mol 25.0g FeO x 1 mol FeO = mol FeO 71.85g FeO Calculating number of moles of Fe, (relate mole ratios in the equation) 0.348mol FeO x 3mol Fe = mol Fe 3mole FeO Calculating the mass of Fe, 0.348mol Fe x 55.85g Fe = 19.4g Fe 1 mol Fe g Al to mol Al to mol Fe to g Fe Steps: IIa IIb IIIb Mm of Al = 27.98g/mol and Mm of Fe = g/mol 25.0g Al x 1 mol Al = mol Al 27.98g Al Calculating number of moles of Fe,(relate mole ratios in the equation) mol Al x 3 mole Fe = mol Fe 2 mole Al Calculating the mass of Fe, mole Fe x 55.85g Fe = 74.9 g Fe 1 mol Fe FeO is the limiting reactant while mass of Fe yielded was 19.4g. Limiting Reactant Involving Volumes of Gas We assume the temperature and pressure remain constant and use the following approach to determine the limiting reactant. I. Calculating the volume of product that can be produced from first reactant. Recall the coefficient in the balanced equation is in the same ratio as the volume of gases. II. Calculate the volume of product that can be produced from the second reactant. Again, the coefficients in the balanced equation are in the same mole ratio as the volumes of gases. III. State the limiting reactant and the corresponding volume of product. The limiting reactant is the gas that gives the least amount of product. The actual volume of product obtained from the

6 chemical reaction is the lesser of two volumes calculated in steps I and II. Example: Suppose 5.00L of NO gas reacts with 5.00L of oxygen gas, which is the limiting reactant and what is the volume of product produced? 2NO (g) + O 2(g) 2NO 2 (g) Step I: 5.00L NO x 2L NO 2 = 5.00L NO 2 2L NO Step II: 5.00L O 2 x 2L NO 2 = L NO 2 1L O 2 Step III: The limiting reactant is NO and the volume of NO 2 produced was 5.00L. PERCENTAGE YIELD Percentage yield = actual yield x 100 Theoretical yield The theoretical yield is the calculated yield of the product while the actual yield is the result you obtained in the lab by going through the procedures required for the product formation. In an experiment, some errors lead to high results while other errors lead to low results. High results will be obtained when impurities are trapped or when the precipitates are not completely dry. Example: 15.0kg of ammonia gives an actual yield of 65.3kg of ammonium nitrate, the calculated yield of ammonium nitrate for experiment is 70.5kg what is the percentage yield? % yield = actual yield x 100 theoretical yield = 65.3kg x kg = 92.6%