# 1.24 Calculations and Chemical Reactions

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1 1.24 Calculations and Chemical Reactions Converting quantities between different substances using a balanced equation A balanced chemical equation tells us the number of particles of a substance which react with another substance. It also therefore tells us the ratios in moles in which these substances react. Eg 2H 2 + O 2 2 H 2 O 2 moles H 2 1 mole O 2 2 moles H 2 O All calculations involving chemical reactions will need quantities given to be converted into amount in moles. Then the balanced equation can be used to identify reacting mole ratios N 2 + 3H 2 2NH 3 The balancing (stoichiometric) numbers are the mole ratios e.g. 1 mol of N 2 reacts with 3 mol of H 2 to produce 2 mol of NH 3 Most calculations will only involve two substances from the reaction and it is only necessary to identify the mole reacting ratio between those two substances. So if a reaction involved 0.5 mol of N 2 we can say it would react with 1.5 mol of H 2 using the 1 N 2 :3 H 2 ratio Or if 0.4 moles of NH 3 was produced we would need 0.6 mol of H 2 using the 2 NH 3 :3 H 2 ratio N Goalby chemrevise.org 1

2 Reacting Mass Questions When doing calculations involving masses of different substances in a chemical reactions, to convert from mass of A to mass of substance B we must use moles. A+B products A+ other reactant B A & B can be any two reactants or products Mass of A Mass of B Step 1. Turn mass of A into moles of A Moles of A It is not possible to go directly from mass of one substance to another. We must use moles Step 2. Use balanced equation to work out moles B from moles of A Moles of B Step 3. Turn moles of B into mass of B Example 1 What mass of Carbon dioxide would be produced from heating 5.5 g of sodium hydrogencarbonate? 2NaHCO 3 Na 2 CO 3 + CO 2 + H 2 O Step 1: work out amount, in mol, of sodium hydrogencarbonate amount = mass / Mr = 5.5 /84.0 = mol Step 2: use balanced equation to give amount in mol of CO 2 2 moles NaHCO 3 : 1 moles CO 2 So HNO 3 : mol CO 2 Remember stoichiometric balancing numbers are not used to calculate Mr. The Mr of NaHCO 3 is always 84.0 whatever the balancing number in front of it Step 3: work out mass of CO 2 Mass = amount x Mr = x 44.0 =1.44g Reacting Mass Questions 1.1) What mass of TiO 2 can be made from reacting 10.0g of TiCl 4 with water? TiCl 4 (l) + 2H 2 O(l) TiO 2 (s) + 4HCl(aq) 1.2) What mass of calcium chloride can be formed when g of calcium reacts with hydrochloric acid? Ca + 2 HCl CaCl 2 + H 2 1.3) What mass of carbon monoxide is needed to reduce 3.00 kg of iron oxide to iron? Fe 2 O CO 2 Fe + 3 CO 2 1.4) What mass of F 2 would be needed to produce 150 g of C 4 F 10? C 4 H F 2 C 4 F HF 1.5) What mass of NO would be produced from g of CuS reacting with concentrated nitric acid? 3 CuS + 8 HNO 3 3 CuSO 4 + 8NO + 4H 2 O N Goalby chemrevise.org 2

3 Limiting and excess reactants If a question gives masses of two of more reactants we will need to consider whether one is in excess. In most cases when there are two reactants, after they have reacted we will be left with some of one reactant. This is the amount that is in excess. Students often find this a complex idea to understand. Let us consider the reaction of A and B reacting with a ration of 1A:2B A+ 2B C If we had 6 moles of A and 10 of B, which is in excess and how much C would be produced? +? We could make 5 moles of C and we would have 1 mole of A left over as excess. B is what we call the limiting reactant So if given masses of two reactants then it is necessary to work out moles of both substances and do the above exercise to work out which one is limited and which one is in excess. Example 2 What is the maximum mass of Titanium that could be produced from reacting 100 g of TiCl 4 with 80 g of sodium TiCl Na 4 NaCl + Ti Step 1: work out amount, in mol, TiCl 4 amount = mass / Mr = 100 /189.9 = mol Step 2 use balanced equation to work out which reactant is in excess Using 1TiCl 4 :4 Na ratio we can see that 0.527mol of TiCl 4 should react with mol of Na. We actually have 3.48 mole of Na which is an excess of moles. We can complete calculation using the limiting reactant of TiCl 4 Step 3: use balanced equation to work out amount in mol of Ti formed 1 mol TiCl 4 : 1 mole Ti So 0.527mol TiCl 4 produces mole Ti Step 4: work out mass of Ti formed Mass = amount x Mr = x 47.9 =25.24g Step 1: work out amount, in mol, Na amount = mass / Mr = 80/23.0 = 3.48 mol 2.1) What is the maximum mass of iron sulphide that can be produced from 20.0 g of iron and 30.0g of sulphur? Fe + S FeS 2.2) What is the maximum mass of ammonia that can be produced from 400 g of ammonium chloride and 100g of calcium oxide? 2 NH 4 Cl + CaO CaCl 2 + H 2 O + 2 NH 3 2.3) What is the maximum mass of aluminium chloride that can be produced from 25.0 g of aluminium and 50.0 g of HCl? 2 Al + 6 HCl 2 AlCl H 2 2.4) What is the maximum mass of iron (III) chloride that can be produced from 50.0 g of Fe and 100g of chlorine gas? 2 Fe + 3 Cl 2 2 FeCl 3 2.5) What is the maximum mass of phosphorus that can be produced from reacting 100g of Ca 5 F(PO 4 ) 3, 100g of SiO 2 and 100g of Carbon? 2Ca 5 F(PO 4 ) 3 + 9SiO C 9CaSiO 3 + CaF CO + 6 P N Goalby chemrevise.org 3

4 Calculations involving masses, solutions and gases Commonly in questions converting between quantities of substances reacting we will use more than just mass data. We might have the volume and concentration of a solution, or the volume of a gas. We need to adapt our existing method or reacting masses to include other quantities. Any of the equations below can be used to convert quantities into moles. 1. For pure solids, liquids and gases 2. For Gases 3. For solutions amount = mass MolarMass Unit of Mass: grams Unit of amount : mol General method Step 1: Use one of the above equations to convert any given quantity into amount in mol Mass amount Volume of gas amount Conc and vol of solution amount Gas Volume (dm 3 )= amount x 24 This equation give the volume of a gas at room pressure (1atm) and room temperature 25 o C. Or use the ideal gas equation to work out gas volumes at other temperatures and pressures PV = nrt Step 2: Use balanced equation to convert amount in mol of initial substance into amount in mol of second substance Concentration = amount volume Unit of concentration: mol dm -3 or M Unit of Volume: dm 3 Step 3 Convert amount, in mol, of second substance into quantity question asked for using relevant equation e.g. amount,mr mass Amount gas vol gas amount, vol solution conc Example 3: What mass of Copper would react completely with 150 cm 3 of 1.6M nitric acid? 3Cu + 8HNO 3 3Cu(NO 3 ) 2 + 2NO + 4H 2 O Step 1: work out moles of nitric acid amount = conc x vol = 1.6 x 0.15 = 0.24 mol Step 2: use balanced equation to give moles of Cu 8 moles HNO 3 : 3 moles Cu So 0.24 HNO 3 : 0.09 (0.24 x 3 / 8 ) mol Cu Step 3: work out mass of Cu Mass = amount x Mr = 0.09 x 63.5 =5.71g Example 4: What is the total volume of gas produced in dm 3 at 333K and 100kPa when g of magnesium nitrate decomposes when heated? 2Mg (NO 3 ) 2 (s) 2 MgO (s) + 4NO 2 (g) + O 2 (g) Step 1: work out moles of magnesium nitrate Moles = mass / Mr = / = mol Step 2: use balanced equation to give moles of gas produced 2 moles Mg (NO 3 ) 2 : 4NO 2 (g) + O 2 (g) ie 5moles of gas So Mg (NO 3 ) 2 : ( x 5 / 2 ) moles gas Step 3: work out volume of gas Volume = nrt/p = ( x 8.31 x 333 )/ = m 3 = 0.303dm 3 N Goalby chemrevise.org 4

5 Questions involving masses, solutions and gases 3.1) Hydrogen can be made by the reaction of hydrochloric acid with magnesium according to the equation 2HCl + Mg MgCl 2 + H 2 What mass of hydrogen is formed when 200cm 3 of hydrochloric acid of concentration 2.5 moldm 3 reacts with an excess of magnesium? 3.2) Lead(II) nitrate can be produced by the reaction between nitric acid and lead(ii) oxide as shown by the equation below. PbO + 2HNO 3 Pb(NO 3 ) 2 + H 2 O An excess of lead(ii) oxide reacted with 250cm 3 of 1.25 moldm 3 nitric acid. Calculate the maximum mass of lead(ii) nitrate which could have been obtained. 3.3) Calculate the volume, in cm 3, of 1.50 moldm 3 hydrochloric acid required to react completely with 1.20 g of magnesium hydroxide. Mg(OH) 2 (s) + 2HCl(aq) MgCl 2 (aq) + 2H 2 O(l) 3.4) A volume of 150cm 3 of 1.75 moldm 3 nitric acid was completely reacted with copper metal. The equation for the reaction is shown below. Calculate the mass of copper that would react completely with this amount of nitric acid. 3Cu + 8HNO 3 3Cu(NO 3 ) 2 + 2NO + 4H 2 O 3.5) A sample of Sodium hydrogencarbonate was heated until completely decomposed. The CO 2 formed in the reaction occupied a volume of 352 cm 3 at Pa and 298 K. Calculate the mass of the NaHCO 3 that has decomposed. 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) 3.6) Oxygen may be prepared by the decomposition of hydrogen peroxide, H 2 O 2, as shown in the equation. 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) A 160 cm 3 sample of 2.68 mol dm 3 aqueous hydrogen peroxide was decomposed completely. Calculate mass and volume of oxygen gas produced at room temperature and pressure. 3.7) Phosphine, PH 3, and oxygen can react to form phosphoric acid, H 3 PO 4, as shown in the equation below. PH 3 + 2O 2 H 3 PO 4 An excess of oxygen was mixed with 2.70 g of phosphine in a sealed container and allowed to react. Calculate the mass of phosphoric acid formed in this reaction. 3.8) Ammonium sulfate reacts with sodium hydroxide to form ammonia, sodium sulfate and water as shown in the equation below. (NH4) 2 SO 4 (s) + 2NaOH(aq) 2NH 3 (g) + Na 2 SO 4 (aq) + 2H 2 O(l) A 3.24 g sample of ammonium sulfate reacted completely with 39.70cm 3 of a sodium hydroxide solution. Calculate the concentration, in moldm 3, of the sodium hydroxide solution used. N Goalby chemrevise.org 5

6 Questions involving samples being taken There are many variations on these calculations. It is not possible to learn an exact method for every type of calculation you could ever meet. You need to be able to go back to the chemistry and picture what is happening. Sometimes it is actually a good idea to draw out what is happening. Let s look at some common extra steps Samples of solutions taken Often in the laboratory we might prepare a solution of a reactant in a volumetric flask which we then want to react with another substance. Rather than reacting all the substance at once we often take out a sample to react. We need to take that into account when doing the calculation. This can be done very simply. 25 cm 3 volumetric pipette 250 cm 3 volumetric flask containing M + ions If the titration tells us there we moles of M + in the conical flask then there must have been x10 in the original sample (0.132) because we removed a tenth of the solution with the pipette conical flask Example 5. An unknown metal carbonate reacts with hydrochloric acid according to the following equation. M 2 CO 3 (aq) + 2HCl(aq) 2MCl(aq) + CO 2 (g) + H 2 O(l) A 3.96 g sample of M 2 CO 3 was dissolved in distilled water to make 250 cm 3 of solution. A 25.0 cm 3 portion of this solution required 32.8 cm 3 of mol dm 3 hydrochloric acid for complete reaction. Calculate the Mr of M 2 CO 3 and identify the metal M 1. Calculate the number of moles of HCl used amount = conc x vol = x = mol 2. Work out number of moles of M 2 CO 3 in 25.0 cm 3 put in conical flask use balanced equation to give moles of M 2 CO 3 2 mol HCl : 1 mol M 2 CO 3 So NaOH : moles M 2 CO 3 3. Calculate the number of moles M 2 CO 3 acid in original 250 cm 3 of solution Moles in 250cm 3 = x10 = work out the Mr of M 2 CO 3 Mr= mass / amount = 3.96/ = Work out Ar of M = ( x3) 2 Ar of M = 39 M= potassium N Goalby chemrevise.org 6

8 Calculations involving two or more reactions One type of question is where there is a series of reactions when one occurs after the other. The product of the first reaction becomes the reactant of the second reaction. 2 Mg(NO 3 ) 2 2MgO + 4NO 2 + O 2 If there was 0.1mol of Mg(NO 3 ) 2 then it would produce 0.1 mol of MgO in the first reaction (Using the 2 Mg(NO 3 ) 2 : 2 MgO ratio) MgO + 2 HCl MgCl 2 + H 2 O Then all the 0.1 mol of MgO made in the first reaction would react with 0.2 mol of MgCl 2 in the second reaction (Using the 1 MgO: 2HCl ratio) Remember the balancing numbers give the mole ratios for that equation. Don t try to link balancing numbers in one reaction to the balancing numbers in the second 5.1) A 2.23g sample of magnesium nitrate was fully decomposed. 2 Mg(NO 3 ) 2 2MgO + 4NO 2 + O 2 The magnesium oxide, produced was reacted with hydrochloric acid. MgO + 2HCl MgCl 2 + H 2 O This sample of magnesium oxide required 33.2cm 3 of hydrochloric acid for complete reaction. Calculate the concentration, in mol dm 3, of the hydrochloric acid. 5.2) Sodium carbonate is manufactured in a two-stage process as shown by the equations below. NaCl + NH 3 + CO 2 + H 2 O NaHCO 3 + NH 4 Cl 2NaHCO 3 Na 2 CO 3 + H 2 O + CO 2 Calculate the maximum mass of sodium carbonate which could be obtained from 600 g of sodium chloride. 5.3) The chloride of an element X reacts with water according to the following equation. XCl 4 (l) + 2H 2 O(l) XO 2 (s) + 4HCl(aq) A g sample of XCl 4 was added to water. The solid XO 2 was removed by filtration and the resulting solution was made up to 250 cm 3 in a volumetric flask. A 25.0 cm 3 portion of this solution was titrated against a mol dm 3 solution of sodium hydroxide, of which 22.3 cm 3 were required to reach the end point. NaOH + HCl NaCl + H 2 O Calculate the relative molecular mass, Mr, of XCl 4 and then deduce the relative atomic mass, Ar, of element X and its identity. 5.4) A old bronze coin of mass 2.15g containing copper and tin was dissolved in concentrated nitric acid and the NO gas escaped through boiling. The equation below shows the reaction of the copper with the acid 3 Cu (s) + 8 HNO 3 (aq) 3 Cu(NO 3 ) 2 (aq) + 2NO (g) + 4H 2 O (l) The solution was diluted with water to make up to 250cm 3. 25cm 3 portions were neutralised and added to excess potassium iodide solution. Only the copper ions reacted with the iodide ions to produce iodine. Equation for reaction 2Cu 2+ (aq) + 4I (aq) 2CuI (s) + I 2 (aq) The liberated iodine was then titrated with sodium thiosulphate. The iodine required 30.40cm 3 of mol dm -3 sodium thiosulphate (Na 2 S 2 O 3 ) to react. I 2 (aq) + 2S 2 O 3 2 (aq) 2I (aq) + S 4 O 6 2 (aq) What is the percentage of Copper in the brass coin? 5.5) The amount of ozone in the atmosphere may be determined by passing air through a solution of acidified potassium iodide to form Iodine in the following reaction O 3 + 2I + 2H + O 2 + H 2 O + I 2 The amount of iodine formed can be determined by titration with a solution of sodium thiosulfate of known concentration in the following reaction I 2 + 2S 2 O 3 2 2I + S 4 O 6 2 In an experiment to determine the amount of ozone in air, 100 m 3 of air was bubbled through 100 cm 3 of a solution containing an excess of acidified potassium iodide. The resulting solution was titrated against a solution of sodium thiosulfate of concentration mol dm 3. The volume of sodium thiosulfate solution required for the complete reaction was cm 3. Calculate the amount in moles of Ozone in the 100m 3 sample and calculate the volume of Ozone in m 3. (assuming the volume is measured at room temperature and pressure). What percentage by volume of air is ozone? N Goalby chemrevise.org 8

9 The second type of calculation involving two different reactions occurs when the first reaction leaves an excess reactant. The second reaction then involves the reaction of the excess reactant. Think about the following series of reactions with A reacting with excess B in a first reaction. Then the remaining B reacts in a second reaction with reactant D. (equations below) If we had 4 moles of A and 12 of B, which is in excess and how much B would be left over? Reaction 1 A+ 2B C + + The 4 moles of B left over would react with 4 moles of D Reaction 2 B+ D E 4 moles of B left over + Now try this problem. If 17 moles of B reacted with a unknown quantity of A and the unreacted B reacted with 3 moles of D. How many moles of A would there have been? The answer is 7 This logic is used to solve a type of calculation we call a back titration Example 6: 950 mg of impure calcium carbonate tablet was crushed cm 3 of 1.00 mol dm 3 hydrochloric acid, an excess, was then added and the mixture was transferred to a volumetric flask. The volume was made up to exactly 100 cm 3 with distilled water cm 3 of this solution was titrated with 11.1cm 3 of mol dm 3 sodium hydroxide solution. What is the percentage of CaCO 3 by mass in the tablet 950 mg of impure calcium carbonate 10cm 3 volumetric pipette transfers 10cm 3 of solution containing unreacted HCl 11.1cm 3 of mol dm 3 sodium hydroxide 50.0 cm 3 of 1.00 mol dm 3 HCl This reaction occurs CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) All the CaCO 3 is reacted some HCl is left over 100 cm 3 volumetric flask containing unreacted HCl made up to 100cm 3 distilled water. conical flask This reaction occurs NaOH +HCl(aq) NaCl(aq) + H 2 O(l) The NaOH reacts with left over HCl in 10cm 3 solution The calculation works backwards using the titration results first. N Goalby chemrevise.org 9

10 950 mg of impure calcium carbonate tablet was crushed cm 3 of 1.00 mol dm 3 hydrochloric acid, an excess, was then added and the mixture was transferred to a volumetric flask. The volume was made up to exactly 100 cm 3 with distilled water cm 3 of this solution was titrated with 11.1cm 3 of mol dm 3 sodium hydroxide solution. What is the percentage of CaCO 3 by mass in the tablet? 1. Calculate the amount in moles of sodium hydroxide used. amount = conc x vol = 0.30 x = mol 2. Work out the amount in moles of hydrochloric acid that reacted with the NaOH NaOH +HCl(aq) NaCl(aq) + H 2 O(l) use balanced equation to give moles of HCl 1 mol NaOH : 1 mol HCl So mol NaOH : mol HCl This is the amount of moles of HCl that were in the conical flask and hence in the 10cm 3 pipette 3. Calculate the amount in moles of hydrochloric acid left in 100 cm 3 of solution. Moles in 100cm 3 = x10 =0.0333mol The 10cm 3 pipetted sample came from the 100cm 3 volumetric flask so to work out the amount of moles in the flask we multiply by Calculate the amount in moles of HCl that reacted with the indigestion tablet. In original HCl 50.0 cm 3 of 1.00 mol dm 3 there is 0.05mol moles of HCl that reacted with the indigestion tablet. = =0.0167mol The amount of moles of HCl that reacted the CaCO 3 is the difference between the total amount of moles of HCl at the start and the moles of HCl left over at the end of this reaction. 5 Use balanced equation to give moles of CaCO 3 that reacted with the HCl CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) 2 mol HCl : 1 mol CaCO 3 So HCl : moles CaCO 3 6. work out the mass of CaCO 3 in original tablet mass= amount x Mr = x = g 7. Calculate percentage of CaCO 3 by mass in the tablet = 0.836/0.950 x100 = 88.0 % Divide the mass of calcium carbonate that reacted by the mass of the original impure tablet X100 N Goalby chemrevise.org 10

11 Back Titration questions 6.1) A g of an impure sample of quicklime was dissolved in 50.0 cm 3 of hydrochloric acid, concentration mol dm 3. The excess hydrochloric acid was titrated with sodium hydroxide solution, concentration mol dm 3, and 17.7 cm 3 was needed to just neutralize the acid. The equations for the reactions involved are shown below. CaO(s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Calculate the percentage purity of the Calcium oxide. 6.2) The amount of sulphur dioxide in the air can be measured by bubbling a known volume of air through iodine solution. Sulphur dioxide converts the iodine to iodide ions according to the following equation. SO 2 (g) + I 2 (aq) + 2H 2 O(l) SO 2 ( 4 aq) + 4H + (aq) + 2I (aq) In an experiment, 75 m 3 of air were passed through 50 cm 3 of iodine, of concentration mol dm 3. The remaining iodine was titrated with sodium thiosulphate solution and reacted with cm 3 of sodium thiosulphate, concentration mol dm 3. I 2 (aq) + 2S 2 O 2 3 (aq) 2I (aq) + S 4 O 2 6 (aq) Calculate the moles of sulphur dioxide that were in the original 75m 3 sample. 6.3) 9.8 g of a indigestion remedy containing calcium carbonate was reacted with 100 cm 3 of 2.00 mol dm -3 hydrochloric acid. (Only the calcium carbonate in the remedy reacts). The resulting solution was made up to 250 cm 3 with water in a volumetric flask. A 25.0 cm 3 portion of this solution required 28.9 cm 3 of mol dm -3 sodium hydroxide for neutralisation. Find the percentage by mass of the calcium carbonate in the indigestion remedy. The equations for the reactions involved are shown below. CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 6.4) 1.13 g of an impure sample of calcium hydroxide was dissolved in 50.0 cm 3 of hydrochloric acid, concentration 1.00 mol dm 3. The resulting solution was made up to 250 cm 3 with water in a volumetric flask. A 25.0 cm 3 portion of this solution required 30.7 cm 3 of mol dm -3 sodium hydroxide for neutralisation. The equations for the reactions involved are shown below. Ca(OH) 2 (s) + 2HCl(aq) CaCl 2 (aq) + 2H 2 O(l) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Calculate the percentage purity of the Calcium hydroxide. 6.5) 1.50g of Bordeaux mixture containing a mixture of calcium hydroxide (Ca(OH) 2 ) and copper(ii) sulfate- 5-water (CuSO 4.5H 2 O) was added to 25.0 cm 3 of hydrochloric acid, concentration 2.00 mol dm 3. The calcium hydroxide in the mixture will neutralise some of the acid. The resulting solution was made up to 250 cm 3 with water in a volumetric flask. A 25.0 cm 3 portion of this solution required 26.7 cm 3 of mol dm -3 sodium hydroxide for neutralisation. Calculate the percentage, by mass, of calcium hydroxide in the sample of Bordeaux mixture. 6.6) Dolomite is a carbonate-containing mineral with formula CaX(CO 3 ) 2 where X is a metal ion. The carbonate ions in Dolomite react with acid. The ionic equation for the reaction between hydrogen ions and carbonate ions is shown. 2H + (aq) + CO 3 2 (aq) CO 2 (g) + H 2 O(l ) A 6.00 g sample of dolomite is dissolved in 35.0 cm 3 of 5.00 mol dm 3 hydrochloric acid, which is an excess. The resulting solution is made up to 100 cm 3 in a volumetric flask, using distilled water cm 3 portions of this solution are titrated against sodium hydroxide cm 3 of mol dm 3 NaOH is required for neutralisation. Calculate the relative formula mass of CaX(CO 3 ) 2 and hence the relative atomic mass and identity of the metal ion X N Goalby chemrevise.org 11

12 % Yield actual yield percentage yield = x 100 theoretical yield % yield in a process can be lowered through incomplete reactions, side reactions, losses during transfers of substances, losses during purification stages. Example 7: 25g of Fe 2 O 3 was reacted and it produced 10g of Fe. What is the percentage yield? Fe 2 O 3 + 3CO 2Fe + 3 CO 2 First calculate maximum mass of Fe that could be produced Step 1: work out amount in mol of Iron oxide amount = mass / Mr =25 / = mol Step 2: use balanced equation to give moles of Fe 1 moles Fe 2 O 3 : 2 moles Fe So Fe 2 O 3 : 0.313moles Fe Step 3: work out mass of Fe Mass = amount x Mr = x 55.8 =17.48g % yield = (actual yield/theoretical yield) x 100 = (10/ 17.48) x 100 =57.2% 7.1) 10.0 g of CaCO 3 of was reacted with H 2 SO 3. The reaction was incomplete and only 10.7g of CaSO 3 was formed. What is the percentage yield of the reaction? H 2 SO 3 + CaCO 3 CaSO 3 + H 2 O + CO 2 7.2) 5.70g of Copper metal was reacted with HNO 3. The reaction was incomplete and only 14.80g of Cu(NO 3 ) 2 was formed. What is the percentage yield of the reaction? 3Cu + 8HNO 3 3Cu(NO 3 ) 2 + 2NO + 4H 2 O 7.3) 5.78 g of magnesium carbonate were added to an excess of sulfuric acid. The following reaction occurred. MgCO 3 + H 2 SO 4 MgSO 4 + CO 2 + H 2 O Calculate the actual mass of MgSO 4 produced in this reaction assuming a 85% yield. 7.4) The reaction from ethanoic acid (CH 3 COOH) has a percentage yield of 82%. What is the minimum mass of butane that would be needed to actually produce 10.2g of CH 3 COOH 2C 4 H O 2 4 CH 3 COOH + 2 H 2 O 7.5) Titanium can be produced from TiO 2 from the following two reactions TiO 2 + C + 2Cl 2 TiCl 4 + CO 2 TiCl 4 + 4Na 4NaCl + Ti The first reaction has a percentage yield of 90% The second reaction has a percentage yield of 95% What is the actual mass of Titanium that can be produced from 1.00kg of TiO 2 taking into account the percentage yields? N Goalby chemrevise.org 12

13 % Atom Economy percentage atom economy = Mass of useful products Mass of all reactants x 100 Do take into account balancing numbers when working out % atom economy. Example 8 : What is the % atom economy for the following reaction where Fe is the desired product assuming the reaction goes to completion? Fe 2 O 3 + 3CO 2Fe + 3 CO 2 % atom economy = (2 x 55.8) (2 x x16) + 3 x (12+16) x 100 =45.8% Sustainable chemistry requires chemists to design processes with high atom economy that minimise production of waste products. Reactions where there is only one product where all atoms are used making product are ideal and have 100% atom economy. e.g. CH 2 =CH 2 + H 2 CH 3 CH 3 If a process does have a side, waste product the economics of the process can be improved by selling the bi-product for other uses 8.1) Calculate the atom economy to make urea from the following equation 2NH 3 + CO 2 (NH 2 ) 2 CO + H 2 O Urea 8.2) Calculate the atom economy to make epoxyethane from the following equation C 2 H 4 + Cl 2 + Ca(OH) 2 C 2 H 4 O + CaCl 2 + H 2 O Epoxyethane 8.3) Calculate the atom economy to make hydrogen from the reaction of zinc with HCl. Zn + 2 HCl ZnCl 2 + H 2 8.4) Calculate the % atom economy for the formation of CH 2 Cl 2 in this reaction CH 4 + 2Cl 2 CH 2 Cl 2 + 2HCl 8.5) Calculate the % atom economy for the formation of C 6 H 12 Cl 2 in this reaction C 6 H Cl 2 C 6 H 12 Cl 2 + 2HCl 8.6) Calculate the % atom economy for the formation of Br 2 in this reaction 2KBr + Cl 2 Br 2 + 2KCl 8.7) Calculate the % atom economy for the formation of NH 3 in this reaction (NH 4 ) 2 SO 4 + 2NaOH 2NH 3 + Na 2 SO 4 +2H 2 O 8.8) Calculate the % atom economy to make sodium carbonate from this reaction CaCO 3 + 2NaCl CaCl 2 + Na 2 CO 3 8.9) Calculate the % atom economy for the formation of TiO 2 in this reaction TiCl 4 (l) + 2H 2 O(l) TiO 2 (s) + 4HCl(aq) N Goalby chemrevise.org 13

14 Answers 1.1) 4.21g 1.2) 1.39g 1.3) 1579g 1.4) 239g 1.5) 0.418g 2.1) 31.5g 2.2) 60.6g 2.3) 61.0g 2.4) 145g 2.5) 18.4g 3.1) 0.500g 3.2) 51.8g 3.3) 27.4cm 3 3.4) 6.25g 3.5) 2.38g 3.6) 6.86g 5.14dm 3 3.7) 7.78g 3.8) 1.24 mol dm ) ) Mr =124 x=1 4.3) ) 18% 4.5) 138 M=39, K 4.6) 12.2 mol dm ) ) ) 72.6, Ge 5.4) 89.8% 5.5) 5.13 x10-6 mol and % (if using ideal gas equation) 6.1) 68.6% 6.2) 2.65 x ) 72.6% 6.4) 63.3% 6.5) 57.6% 6.6) Mr184.3,Ar 24.2, Mg 7.1) 89% 7.2) 88% 7.3) 7.02g 7.4) 6.01g 7.5) 513g 8.1) 76.9% 8.2) 25.4% 8.3) 1.44% 8.4) 53.8% 8.5) 68.0% 8.6) 51.7% 8.7) 16% 8.8) 48.8% 8.9) 35.4% N Goalby chemrevise.org 14

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