chemical equations. chemical equations we can describe what happens in a chemical reaction using...

Chemical equations

Chemical Equations• 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 equationsRules:

1. Chemical formulas can’t be changed

2. Chemical formulas can be multiplied by a suitable number

Balance the following equation:Fe + O2 Fe3O4

Iron atoms:Reactant side Product side

1 3

2 4Oxygen atoms:

•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

Step one – see what's present

Step two – make changes

Step three – check its balanced

3Fe + 2O2 Fe3O4

Reactant side Product side

3 3

4 4

Iron atoms:

Oxygen atoms:

Try this one: N2 +O2 NO

Nitrogen atoms:


2 1

2 1Oxygen atoms:



• N2 +O2 2NO


Nitrogen atoms:


2 2

2 2Oxygen atoms:

Try this one: Al + Fe2O3 Al2O3 + Fe

Aluminium atoms:


1 2

2

3

1

3

Iron atoms:




Oxygen atoms:

2Al + Fe2O3 Al2O3 + 2Fe


2 2

2

3

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 Aluminium

1 mole of Iron oxide

1 mole of Aluminium Oxide

2 moles of Iron

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

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:

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?

N2 +3H2 2NH3

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

The balanced equation:

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?

2H2O2 O2 +H2O

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..

Q293

(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


C6H1206 2C2H5OH + 2CO2


Q293

(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


C6H1206 2C2H5OH + 2CO2


Q293

(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


C6H1206 2C2H5OH + 2CO2


Q293

(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

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?

Q294

(i) Find how many moles of calcuim carbonate are reacted ( grams – moles)(ii) Find how many moles of sulfur dioxide would react with this calcuim

carbonate ( use balanced equation)(iii) Find what the volume of sulfur dioxide gas would be reacted (moles –

volume)

1000g/ RMM = moles of calcuim carbonate100/ 100 = 10

10 moles of calcuim carbonate would be used




Q293

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



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 sulfur dioxide would react with this calcuim carbonate

( use balanced equation)(i) Find what the volume of sulfur dioxide gas would be reacted (moles –

volume)

Q294

• 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



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)

Q294

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

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

.05 moles 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 SO2 x 22.4L = Volume at stp

• (.05)(22.4) = 1.12 • 1.12L of sulfur dioxide would be obtained

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

1144g of carbon dioxide so change to moles BY DIVIDING BY rmm

1144/ 44 = 26

26 moles of carbon dioxide are used every day

(ii) Each day, how many moles of CO2 are released?

• 26 moles of CO2

• (26)(22.4) = 582.4L

• 582L of carbon dioxide would be produced per day


(iii) Each day, what volume of CO2 is 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

2112g of carbon dioxide / RMM = moles

2112/ 44 = 48

48 moles of carbon dioxide are used every day

(ii) Each day, how many moles of CO2 are released?

• (48)(22.4) = 1075.2L

• 1075.2L of carbon dioxide would be produced per day


(iii) Each day, what volume of CO2 is released?





(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.



(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

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

24.657512.3288



(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

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

24.657512.3288

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

1.0mole / 0.0207moles = cm3 of HCl would be needed

What volume of 1.0M HCl would contain 0.020689655 moles?

ii) What mass of salt is formed in this neutralisation?

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

The balanced equation :Mg(OH)2 + 2HCl MgCl2 + 2H20 1 2 1 2

0.0103 0.0103

iii) How many magnesium ions are present in this amount of salt?

How many ions in 0.0103 moles of the salt?

0.0103moles x 6 x 1023 = 6.2068962 x 10 21

For every molecule of MgCl2 there will be one Mg ion.

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 is3Zn + 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

Q299. A mass of 13g of granulated zinc was reacted with 100cm3 of a 2M solution of nitric acid. The equation for the

reaction is3Zn + 8NO3 3Zn(NO3)2 + 2NO + 4H20

(i) 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

x + 4(-2) = -1 x – 8 = -1 x = -1+ 8 x = 7

( + 7 ) ( - 2) ( -1) ( +1) (+2) ( +1 ) ( -2)

•Each Mn goes down 5 in number (reduction)•RIG – Each Mn is gaining 5 electrons.

•Each Cl goes up 1 in number ( oxidation)•OIL – Each Cl is losing 1 electron

(0)

5

Once the oxidation numbers are balanced, Make sure the overall equation still balances...

2.5 48

Ratio has to be 1 Mn : 5 Cl

Cr2O7―2+ Fe+2 + H+ Cr+3 + Fe+3

+ H2O

x + 7(-2) = -2 2(x) – 14 = -2 2x = 12 x = 6

( + 6 ) ( - 2) ( +2) ( +1) (+3) ( +1 ) ( -2)

•Each Cr goes down 3 in number (reduction)•RIG – Each Cr is gaining 3 electrons.

•Each Fe goes up 1 in number ( oxidation)•OIL – Each Fe is losing 1 electron

(+3)

6

Once the oxidation numbers are balanced, Make sure the overall equation still balances...

2 6 14

Ratio has to be 1 Cr : 3 Fe

7

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