Chemical equations

Chemical Equations Carbon + oxygen Carbon dioxide We can describe what happens in a chemical reaction using words: Carbon + oxygen Carbon dioxide “+” means “and” means “react to give”

Chemical Equations Another way in which we can we can describe a chemical reaction is using symbols to represent the reactants and products instead of words! C + O CO2

Law of conservation of Mass The law of conservation of matter states that, in any chemical reaction, matter doesn’t get created or destroyed but it only changes from one form into another. C + O CO2 In a chemical reaction: No. of atoms of an element present at the start of reaction = No. of atoms of the element present at the end of the reaction

Balanced Chemical Equations Chemical equations should always be balanced to describe accurately what happens during a chemical reaction C + O CO2 For an equation to be balanced the total number of atoms of each element of reactant must equal the total number of atoms of that element in the product

Balancing chemical equations Rules: Chemical formulas can’t be changed Chemical formulas can be multiplied by a suitable number

Balance the following equation: Fe + O2 Fe3O4 Step one – see what's present Reactant side Product side 1 3 2 4 Iron atoms: Oxygen atoms: Step two – make changes To get 3 Fe atoms on the reactant side multiply it by 3! To get 4 oxygen atoms on the product side multiply it by 2! 3Fe + 2O2 Fe3O4

3Fe + 2O2 Fe3O4 3 4 Reactant side Product side Step three – check its balanced 3Fe + 2O2 Fe3O4 Reactant side Product side 3 4 Iron atoms: Oxygen atoms:

Try this one: N2 +O2 NO N2 +O2 2NO 2 1 2 Reactant side Product side Step one – see what's present Reactant side Product side 2 1 Nitrogen atoms: Oxygen atoms: Step two – make changes N2 +O2 2NO Step three – check its balanced Reactant side Product side 2 Nitrogen atoms: Oxygen atoms:

2Al + Fe2O3 Al2O3 + 2Fe Try this one: Al + Fe2O3 Al2O3 + Fe 1 2 3 Step one – see what's present Reactant side Product side 1 2 3 Aluminium atoms: Iron atoms: Oxygen atoms: Step two – make changes 2Al + Fe2O3 Al2O3 + 2Fe Step three – check its balanced Reactant side Product side 2 3 Aluminium atoms: Iron atoms: Oxygen atoms:

Calculations based on balanced chemical equations A balanced equation tells you the relative amounts of each reactant and each product involved in the reaction. In this reaction: 2Al + Fe2O3 Al2O3 + 2Fe 2 moles of Iron 2 moles of Aluminium 1 mole of Iron oxide 1 mole of Aluminium Oxide

Calculations based on balanced chemical equations The reaction between oxygen and nitrogen is described by the balanced chemical equation: Question: If 2 moles of nitrogen were reacted: (i) How many moles of O2 would it react with? (ii) How many moles of NO would be formed? N2 +O2 2NO

Answer… N2 +O2 NO Moles in B.E: 1 1 2 therefore… 2 2 4 Two moles of O2 would react Four moles of NO would be produced

Answer: N2 +3H2 2NH3 The balanced equation: N2 +3H2 2NH3 Question: If 2 moles of nitrogen reacted in this reaction, How many moles of hydrogen would react? How many moles of ammonia would be formed? Answer: N2 +3H2 2NH3 Moles in B.E: 1 3 2 therefore… 2 6 4 Six moles of H2 would react Four moles of NH3 would be produced

The balanced equation: 2H2O2 O2 +H2O Question: If 6 moles of H2O2 reacted in this reaction, How many moles of oxygen would be formed? How many moles of water would be formed? Answer: 2H2O2 O2 +H2O Moles in B.E: 2 1 1 therefore… 6 3 3 Three moles of O2 would be produced Three moles of NH3 would be produced

Q293 The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C6H1206 2C2H5OH + 2CO2 If 126g of glucose are consumed.. (i) How many moles of glucose does this represent? 126g/ RMM = moles of glucose (126) /180 = 0.7 It represents 0.7 moles of glucose

Q293 The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C6H1206 2C2H5OH + 2CO2 If 126g of glucose are consumed.. (ii) How many moles of ethanol are produced? Answer: C6H1206 2C2H5OH + 2CO2 Moles in B.E: 1 2 2 therefore… 0.7 1.4 1.4 1.4 moles of ethanol would be produced

Q293 The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C6H1206 2C2H5OH + 2CO2 If 126g of glucose are consumed.. (i) What volume of carbon dioxide, measured at s.t.p is produced? 1.4 moles of CO2 x 22.4 = Volume of gas at stp (1.4)(22.4) = 31.36 31.36 L of carbon dioxide will be formed

Q293 The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: C6H1206 2C2H5OH + 2CO2 If 126g of glucose are consumed.. (i) How many molecules of carbon dioxide does this volume contain? 1.4 moles of CO2 x 6 x 10 23 = molecules of carbon dioxide (1.4)(6 x 1023) = 8.4 x 1023 8.4 x 1023 molecules of carbon dioxide would be formed

Q294 The following reaction may be used to reduce emissions of sulfur dioxide in waste gases 2CaCO3 + 2SO2 + O2 2CaSO4 +2CO2 What volume of sulfur dioxide ( measured at S.T.P) could be removed from the waste gases by this reaction for every kilogram of calcuim carbonate used? Find how many moles of calcuim carbonate are reacted ( grams – moles) Find how many moles of sulfur dioxide would react with this calcuim carbonate ( use balanced equation) Find what the volume of sulfur dioxide gas would be reacted (moles – volume) 1000g/ RMM = moles of calcuim carbonate 100/ 100 = 10 10 moles of calcuim carbonate would be used

Q294 Q293 The following reaction may be used to reduce emissions of sulfur dioxide in waste gases 2CaCO3 + 2SO2 + O2 2CaSO4 +2CO2 What volume of sulfur dioxide ( measured at S.T.P) could be removed from the waste gases by this reaction for every kilogram of calcuim carbonate used? The fermentation of glucose results in the formation of ethanol and carbon dioxide according to the equation: 2CaCO3 + 2SO2 + O2 2CaSO4 +2CO2 If 126g of glucose are consumed.. Find how many moles of sulfur dioxide would react with this calcuim carbonate ( use balanced equation) Find what the volume of sulfur dioxide gas would be reacted (moles – volume) Answer: 2CaCO3 + 2SO2 + O2 2CaSO4 +2CO2 Moles in B.E: 2 2 1 2 2 therefore… 10 10 5 10 10 10 moles of sulfur dioxide would react with this much caluimcarbonate

Q294 The following reaction may be used to reduce emissions of sulfur dioxide in waste gases 2CaCO3 + 2SO2 + O2 2CaSO4 +2CO2 What volume of sulfur dioxide ( measured at S.T.P) could be removed from the waste gases by this reaction for every kilogram of calcuim carbonate used? Find what the volume of sulfur dioxide gas would be reacted (moles – volume) 10 moles of SO2 x 22.4 = Volume of gas at stp (10)(22.4) = 224L 224L of sulfur dioxide would react, so thats how much would be removed by doing this reaction

295. (ii) A solution of sodium hypochlorite NaOCl is labelled as having a concentration of 5% (w/v). Express the concentration in grams per litre. 5%w/v means 5g in 100cm3. How much in a litre? (5/ 100) x 1000 = 50g Answer: There are 50g of NaOCl in a litre

100cm3 of this 5%w/v solution were reacted with excess chloride ion according to the equation OCl- + Cl- + 2H+ Cl2 + H20 (iii) How many molecules of chlorine gas were liberated? 5%w/v sodium hypochlorite solution means 5g in 100cm3. How many moles? 5g / RMM = Moles NaOCl- 5g/ 74.5 g = 0.0671 There are 0.0671 moles of NaOCl that reacted There must be 0.0671 moles of Cl2 that are made in the reaction

(iii) How many molecules of chlorine gas were liberated? 0.0671moles of Cl2 x 6 x 10 23 (0.0671)(6 x 1023) = 1x 4.0268 x 1022 = x 4.0268 x 1022 molecules of carbon dioxide would be formed

(i) How many moles of sodium sulfite were used? 296. Sulfur dioxide was prepared by heating excess dilute hydrochloric acid with 6.3g of sodium sulfite according to the equation: Na2SO3 + 2HCl NaCl + SO2 + H20 (i) How many moles of sodium sulfite were used? 6.3g of sodium sulfite / RMM = Moles 6.3/ 126 = 0.05 0.05 moles of sodium sulfite would be used

(i) What volume of sulfur dioxide would be obtained? 296. Sulfur dioxide was prepared by heating excess dilute hydrochloric acid with 6.3g of sodium sulfite according to the equation: Na2SO3 + 2HCl NaCl + SO2 + H20 (i) What volume of sulfur dioxide would be obtained? Na2SO3 + 2HCl NaCl + SO2 + H20 Moles in B.E: 1 2 1 1 1 therefore… .05 .10 .05 .05 .05 .05 moles of sulfur dioxide would be obtained .05 moles of SO2 x 22.4L = Volume at stp (.05)(22.4) = 1.12 1.12L of sulfur dioxide would be obtained

(ii) Each day, how many moles of CO2 are released? 300. 143g of carbon dioxide for every kilometre travelled Car is used for 8km every day (i) Each day, what mass of CO2 are released? ( 143) x 8 = 1,144g of carbon dioxide used per day (ii) Each day, how many moles of CO2 are released? 1144g of carbon dioxide so change to moles BY DIVIDING BY rmm 1144/ 44 = 26 26 moles of carbon dioxide are used every day

(iii) Each day, what volume of CO2 is released? 300. 143g of carbon dioxide for every kilometre travelled Car is used for 8km every day (iii) Each day, what volume of CO2 is released? 26 moles of CO2 (26)(22.4) = 582.4L 582L of carbon dioxide would be produced per day

(ii) Each day, how many moles of CO2 are released? 300. Large SUV emits 264g of carbon dioxide per kilometre, how many more litres of CO2 would be released into the environment (i) Each day, what mass of CO2 are released? ( 264) x 8 = 2,112g of carbon dioxide used per day (ii) Each day, how many moles of CO2 are released? 2112g of carbon dioxide / RMM = moles 2112/ 44 = 48 48 moles of carbon dioxide are used every day

(iii) Each day, what volume of CO2 is released? 300. 143g of carbon dioxide for every kilometre travelled Car is used for 8km every day 300. Large SUV emits 264g of carbon dioxide per kilometre, how many more litres of CO2 would be released into the environment (iii) Each day, what volume of CO2 is released? (48)(22.4) = 1075.2L 1075.2L of carbon dioxide would be produced per day

300. Large SUV emits 264g of carbon dioxide per kilometre, how many more litres of CO2 would be released into the environment (iii) Each day, how much more Co2 is released in litres when the SUV is used? 1075.2L – 582L = 493.2L more carbon dioxide was released when driving the SUV

297.Bottle of contents 2.5L of concentrated hydrochloric acid were spilled. It was neutralised with sodium carbonate. The spilled acid was 36%(w/v) . (i) Calculate the number of moles of hydrochloric acid spilled. 2.5 L of a 36% (w/v) HCl solution is how many moles? 36/100 x 2500 = 900g There are 900 g of HCL in this much solution. 900g of HCl is how many moles? 900g / RMM = moles 900/ 36.5g = 24.6575 There are 24.6575 moles of HCL in this much solution.

297.Bottle of contents 2.5L of concentrated hydrochloric acid were spilled. It was neutralised with sodium carbonate. The spilled acid was 36%(w/v) . (i) What is the minimum mass of anhydrous sodium carbonate required to completely neutralise the spilled acid?. The balanced equation of the reaction is Na2CO3 + 2HCl 2NaCl + H20 + CO2 1 2 2 1 1 24.6575 12.3288 What is the mass of 12.3288 moles of sodium carbonate? 12.3288 XRMM = moleS 12.3288 X 106 = 1306.8493g 1306.8493g anhydrous sodium carbonate are required to completely neutralise the spilled acid

297.Bottle of contents 2.5L of concentrated hydrochloric acid were spilled. It was neutralised with sodium carbonate. The spilled acid was 36%(w/v) . (i) What volume of carbon dioxide in L ( at STP) would be produced in this neutralisation reaction? The balanced equation of the reaction is Na2CO3 + 2HCl 2NaCl + H20 + CO2 1 2 2 1 1 24.6575 12.3288 What is the volume of 12.3288 moles of carbon dioxide? 12.3287 moles X 24 = Gas volume 12.3287 X 24L = 295.8 L of carbon dioxide gas( at STP) would be produced in this neutralisation reaction

Q298. An indigestion table contains a mass of 0 Q298. An indigestion table contains a mass of 0.30g of magnesium hydroxide. Balanced equation for reaction: Mg(OH)2 + 2HCl MgCl2 + 2H20 (i) Calculate the volume of 1.0M HCl neutralised by two of these digestion tablets. Give your answer to the nearest cm3 Mass of Mg(OH)2 in two digestion tablets = 0.30 x 2 = 0.60g Moles of Mg(OH)2 in 0.60g of tablets 0.6g / RMM = mole 0.6/58 = 0.0103 moles The balanced equation : Mg(OH)2 + 2HCl MgCl2 + 2H20 1 2 1 2 0.0103 0.0207 moles of HCl would be needed

What volume of 1.0M HCl would contain 0.020689655 moles? 1.0mole / 0.0207moles = cm3 of HCl would be needed

ii) What mass of salt is formed in this neutralisation? The balanced equation : Mg(OH)2 + 2HCl MgCl2 + 2H20 1 2 1 2 0.0103 0.0103 What is the mass of 0.0103 moles of the salt? 0.0103 moles /RMM = mass 0.0103/ 95 g = 0.9828g X = 0.9828g of the salt is formed

iii) How many magnesium ions are present in this amount of salt? For every molecule of MgCl2 there will be one Mg ion. How many ions in 0.0103 moles of the salt? 0.0103moles x 6 x 1023 = 6.2068962 x 10 21

Another remedy of Mg(OH)2 in water is marked 6%(w/v) Another remedy of Mg(OH)2 in water is marked 6%(w/v). What volume of this second remedy would have the same effect as the tablets from before? We need to find what volume of the second remedy would contain 0.0103 mole of Mg(OH)2 Remedy has 6g of Mg(OH)2 in 100cm3. How many moles are in 6g? This remedy has 6g in 100cm3 6g/ 58 = 0.1034 moles There are 0.1034 moles of Mg(OH)2 in 100cm3 So 10cm3 OF THE SOLUTION WOULD BE NEEDED

Q299. A mass of 13g of granulated zinc was reacted with 100cm3 of a 2M solution of nitric acid. The equation for the reaction is 3Zn + 8NO3 3Zn(NO3)2 + 2NO + 4H20 (i) Show clearly that the zinc is in excess in the reaction 13g of zinc / RMM = moles of Zinc present

Show clearly that the zinc is in excess in the reaction Q299. A mass of 13g of granulated zinc was reacted with 100cm3 of a 2M solution of nitric acid. The equation for the reaction is 3Zn + 8NO3 3Zn(NO3)2 + 2NO + 4H20 Show clearly that the zinc is in excess in the reaction 2moles/ 1000 X100 = 0.2 moles X = 0.2 moles of NO3 present in the reaction

3Zn + 8NO3 3Zn(NO3)2 + 2NO + 4H20 3 8 3 2 4 So Zinc is in excess in the reaction

What mass of zinc nitrate was formed?

Balancing redox equations

MnO4―+ Cl-1 + H+ Mn+2 + Cl2 + H2O Each Mn goes down 5 in number (reduction) RIG – Each Mn is gaining 5 electrons. Once the oxidation numbers are balanced, Make sure the overall equation still balances... MnO4―+ Cl-1 + H+ Mn+2 + Cl2 + H2O 4 5 8 2.5 ( + 7 ) ( - 2) ( -1) ( +1) (+2) (0) ( +1 ) ( -2) x + 4(-2) = -1 x – 8 = -1 x = -1+ 8 x = 7 Each Cl goes up 1 in number ( oxidation) OIL – Each Cl is losing 1 electron Ratio has to be 1 Mn : 5 Cl

Cr2O7―2+ Fe+2 + H+ Cr+3 + Fe+3 + H2O Each Cr goes down 3 in number (reduction) RIG – Each Cr is gaining 3 electrons. Once the oxidation numbers are balanced, Make sure the overall equation still balances... Cr2O7―2+ Fe+2 + H+ Cr+3 + Fe+3 + H2O 6 14 2 6 7 ( +2) ( +1 ) ( -2) ( + 6 ) ( - 2) ( +1) (+3) (+3) Cr2O7 2— + Fe2+ + H+  Cr3+ + Fe3+ + H2O x + 7(-2) = -2 2(x) – 14 = -2 2x = 12 x = 6 Each Fe goes up 1 in number ( oxidation) OIL – Each Fe is losing 1 electron Ratio has to be 1 Cr : 3 Fe