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### 3. Equations.

posted Mar 18, 2016, 1:53 AM by Upali Salpadoru   [ updated Aug 29, 2019, 9:55 PM ]

Chemical Equations.

An equation balances the reactants and the products formed due to a chemical reaction.

What are the things that have to balance?

. Fig.1.  A balanced equation.

• Number of molecules?

• Left =1.   Right = 2. …..No

• Number of atoms?

•  Left = 5    Right =  3+ 2 …..Yes !

• Total mass ?         Left = 40+12+ 48     , Right =  (40+16) + (12+32)......   Yes!

• Total volume ?    Left=CaCO3 has a very small volume,

Right= As CO2 is a gas , it has a huge volume…….     No.

Balancing chemical equations.

Burning of the carbon compound, Methylene in air is a chemical reaction.

Word equation:-    Methylene + Oxygen = Carbon dioxide + Hydrogen oxide.

Using symbols:-   CH2  +    3O2   → CO2+H2O     …X. ..................This is not balanced.

Subscript 2 gives the number of hydrogen atoms.

3 gives the number of oxygen molecules.

In order to balance the above equation try doubling the blue molecules.

2CH2  +    3O2   → 2CO2+2H2O   He Presto!  You get it right.

What more do the Equations give?

A word equation gives you reactants and the  products of a chemical reaction. A balanced equation in formulae form gives much more.

Sometime we include much more information than just the number of atoms, molecules and their weights.

Examine this:-

2CH4(g). + 3O 2(g) → C(s). +CO2 (g)+4H2O(l).,   Δ Heat.

This shows the incomplete combustion of methane gas. The subscripts letters s, l, and g stand for the state of the substance at room temperature.

(S)- solid. (l)- liquid and (g) for gas. Δ H,  shows that the reaction releases energy.

Sometimes gases that escape and the precipitates are indicated by vertical arrows.

As in this .

•  What happens when a solution of sodium chloride (Common salt) is mixed with   a silver nitrate solution? You get  a white precipitate.

NaCl(aq) + AgNO3 (aq) → NaNO3(aq) + AgCl

What does (aq) mean?

Aqueous is a word for water.

• Burning sulphur produces sulphur dioxide gas.

S(s) + O2(g) →  SO2

Reversible reactions.

Equations also indicate the direction. Always the reactants on the left tend to form the products on the right. But sometimes, certain reactions can work in the reverse direction under certain conditions. These are called reversible reactions. A special two way arrow is used to indicate these.

Example.

1.           NH4Cl(s)   NH3↑  + HCl↑

2.            H2SO4    ⇋   2H+  +   SO42-

Here the sulphuric acid molecule gets split up into its radicals.  Two hydrogen ions each witha positive charge and a sulphate ion with a negative charge is formed.

Example.

Study the equation given here for the heating of condys: Two molecules of the chemical,  2KMnO4 , have 8 atoms. But they release only 2 atoms of oxygen when heated.

Fig.2. Potassium permanganate crystals.

Potassium permanganate  = potassium manganate + Oxygen.

2KMnO4           =  K2MnO4 + MnO2 + O2

.

Example 1

• Find the mass of oxygen that may be obtained from 1.kg of Potassium permanganate

• We can easily solve the problems like this using equations.

 Reactants = Products Equation                        2KMnO4 → K2MnO4   + MnO2 +   O2 Mass. 2x(39.1+54 +64) = 314.2g. → Gives………….32g of Oxygen. 1 kg of KMnO4 → Gives  32 /314.2   of Oxygen.    =  0.1kg of Oxygen.

Example 2

Pure Barium chloride was dissolved in distilled water and excess of dilute sulphuric acid was added.

Equation”   BaCl2  + H2SO4   =  BaSO4 + 2HCl

Mass of BaCl2 taken…………10.4 g

Volume of solution made …………   1000ml.

Volume of solution taken  ………….  500 ml.

Calculate the mass of the precipitate formed.

 Reactants = Products Equation:           BaCl2  + H2SO4 → BaSO4 ↓ + 2HCl Molecular masses:   137+71 BaCl2                                      208             (½ of 10.4)        5.2 g →→→ 137+32.1 +64   BaSO4 ↓   233.1  233.1 ÷ 208x5.2  = 5.83 g,

“Chemical reactions take place in definite ratios”Courtsey Fotofolia.File: #6758598 | Author: Rey Kamensky What happens if you select a wrong ratio?   Let us take an example.

Fig.1 Ball room dancers.

In the figure  you will see a group of 10 ladies and 7 men. Study how they have grouped.

In chemical combinations same thing can take place.

Those poor atoms unable to find partners will stay without combining.    Mehara has taken  30g of iron filings Nuran added 16 g of sulphur.

Let us study the equation for combining Iron and sulphur.

 Reactants = Products Word Equation:     Iron + Sulphur  In symbols: Fe   +    S Formula mass          56   +   32 →→→ Iron sulphide.FeS88

For    56g   of Fe to combine  you need = 32g   of sulphur.

For    1g   of Fe to combine  you need = 32÷ 56   of sulphur.

For  30g                                       = 32÷56 x 30

= 17.1g of sulphur.

So they don't have enough sulphur.

Explanation in another way:-

 The required ratio of Iron:Sulphur = 28: 16Atoms in 2g of Mehara’s iron cannot find partners.16g of sulphur can combine only with 28 g Mass of iron left would be=  30 – 28 = 2g.      2g of iron will not find partners. 1.0   Identify what is shown in RED.  Select the answers from these.

A-.  water solution.  B-. Gas. C- Solid.     D- Liquid.E- Number of molecules.     F- Number of atoms in 1 molecule.  G- Reversible reaction. H-Releasing energy.         I- Charge  J-  Precipitate.

 Question4 Al +3O2 → 2Al2O32Al + 6HCl → 2AlCl3 + 3H2CaCO3(s) →  CaO  + CO2 SO2(g) + H2O(l) →  H2SO3 (l) Mg(s) + CuSO4(aq) → MgSO4 (aq)+ Cu(s)CaCO3 ⇋   CaO  + CO22H2  +  O2 →  2H2O(l)HNO3 ⇋ H+ +  NO3-Ca(OH)2  + CO2 →CaCO3 ↓+ H2O   C (S)  +  O 2(g)    →   CO 2(g)    +ΔH=  393.5kJ Answer.1.F2.E3.C4.B5.A6.G7.D8.I.9.J10.H

2x10 =20 marks

2.0  Write the word equations given here using formulas and balance them.

1. Carbon -C + Oxygen O2 →  Carbon monoxide.- CO

2.  Hydrogen -H2,  + Chlorine- Cl2   →  Hydrogen chloride.-HCl,

3.  Zinc-Zn + Sulphuric acid H2SO4 Zinc sulphate ZnSO4 + hydrogen-H2

4.Sodium carbonate-Na2CO3+Nitric acid HNO3 Sodium nitrate NaNO3  + Carbonic     acid-H2CO3

5. Iron-Fe  +  Oxygen- O2→  Iron oxide - Fe2O3

3x5=15

3.0 Fill in the formula masses.( Molecular mass x number of molecules)

1.                                                    Example.

 Reactants.C     +       O2     = Products.CO2 Reactants.Mg  + Cl2 Products. MgCl2 24  + 71 95

2.

 NH4Cl    = NH3   +  HCl

3.

 CO2  +  Ca(OH)2  = CaCO3 +  H2O

 4. (NH4)2SO4 + CaCO3      = (NH4)2CO3 + CaSO4

 5.     Cu       +       2H2SO4—–> CuSO4  +  SO2  +  2H2O

3x5-15

4.0. Hydrogen is prepared by adding Zinc to dilute hydrochloric acid. Find the mass of the metal required to get 1 gram of Hydrogen.

Zn + 2 HCl   →  ZnCl2 + H2.

5.0 Magnesium reacts with hot water to give hydrogen and Magnesium hydroxide.

Find the mass of the hydrogen formed from 1g of magnesium.

Mg + 2 H2O →  Mg(OH)2 + H2

5 marks

6.0   Petrol has a compound named Octane. This combines with oxygen releasing energy for the internal combustion engine. Find the mass of Carbon dioxide produced and the mass of oxygen consumed for 10 tonnes(t) of octane.

2 C8 H18  +25 O2   16 CO2 + 18H2O

5 marks

7.0 The only ammonium compound in a sample of a particular fertilizer is Ammonium sulphate. If the 200g of sample released 8.5g of ammonia Find the percentage of Ammonium sulphate in the sample.

(NH4)2SO4  → H2SO4 + 2 NH3

6 marks

8.0 Two most important chemical reactions in the world for life are photosynthesis and . Find

(i) the mass of Oxygen produced during photosynthesis giving 100 g of glucose. (ii) The amount of oxygen consumed for 100g of Glucose during respiration..

Photosynthesis...

Carbon Dioxide + Water------->Glucose + Oxygen

6 CO 2 + 6H2O ----->  C 6H 12O 6 +  6O2

2x4=8 marks.

9.0     Amounts of chemicals taken.

Copper  ……64g and Sulphur…….40g.

Equation :             Cu    +  S     --->   CuS

Formula mass 64   +32.          --->   96

i.                   Find the mass of copper sulphide formed.   =

ii.                 Which is in excess?                                      =

iii.              How much of it would be left un-combined?  = .

2x3 =6 mars

10.0       A piece of coal of mass 12g was burnt in a jar containing 16g of oxygen.

(i)  .Which is the proper equation for this combination.  A or B.?

(ii.) Suggest a method for it to combine according to the other equation.

A-    C + O2 --->   CO2.

B-     2C + O2  ---> 2CO.

2x4 =8 marks

11.0.     Acetylene gas under certain conditions may undergo incomplete combustion as follows:Find the mass of unburnt carbon from 1 kg of acetylene.

3C2H2 + 3O2   --->  CO2 +CO + 4C +3H2O

6 marks 12.0   Zinc and sulphur gives an excellent mixture that can burn as follows.

If 6 g of Zn is mixed with 2 gof S find the mass of ZnS formed.

Equation:........Zn  +      S    --->  Zn S

Formula mass ….. 65+    32    =  97

6 marks.

SOLUTIONS

1.0

1.F , 2.E, 3.C,4.B,5.A,6.G,7.D,8.I, 9.J,10.H,

2.0  Write the word equations given here using formulas and balance them.

1.                2C + O2  2 CO

2.                H2   +Cl2   2HCl

3.                         Zn  + H2SO4 ZnSO4 + H2

4.                Na2CO3+ 2HNO3 2NaNO3 + H2CO3

5.                4Fe  +  3O2→   2Fe2O3

3.0 Fill in the formula masses.( Molecular mass x number of molecules)

1.                                                   .

 Reactants.C     +       O2     = Products.CO2 12    +   32 44

2.

 NH4Cl    = NH3   +  HCl 53.5 17       + 36.5

3.

 CO2  +  Ca(OH)2  = CaCO3 +  H2O 44   +  74 100   +18

 4. (NH4)2SO4 + CaCO3      = (NH4)2CO3 + CaSO4 (4x 132)   +100 96   +  136

 5.     Cu       +       2H2SO4—–> CuSO4  +  SO2  +  2H2O 64        +      2x98=196 160      +  64   +36

3x5-15

4.0          Zn + 2 HCl   →  ZnCl2 + H2.

65 …………………….2

2g of H require…….65

1g…………………...65/2 = 32.5g                                                                              5.0    Mg + 2 H2O →  Mg(OH)2 + H2

24g …………………………...2g

1g of Mg gives      …………...2/24  =  1/12g or 0.08g

6.0         2 C8 H18  +25 O2   16 CO2 + 18H2O

2(96+18)  tonnes  give   ,,,,,,,,16x44 t of CO2

228 t  ………………………….704 t.

10 t …………………………..704 ÷228x 10.  =  30.88 t

7.0         (NH4)2SO4  → H2SO4 + 2 NH3

36+32+64……………...34

34 g of NH3 is from     132 g of salt.

1g  …………………….132÷ 34

8.5 g   from……………..132÷ 34  x 8.5 = 33 g

200g of the sample would have contained only 33g of am.sulphate.

As a percentage    33÷ 200x100  = 16.5%.

8.0          6 CO 2 + 6H2O ----->  C 6H 12O 6 +  6O 2

(72+ 12+96)  .+    6 x 32

180 g  of glucose comes with  192g of O2.

1g …………………………….192÷ 180 g.

(i)    100g ……………………..192÷ 180 x100 g  =  106.7g

(ii)   For respiration the same amount has to be used up

9.0          i.                Find the mass of copper sulphide formed.   =96g

ii.                Which is in excess?                                      = Sulphur.

iii.              How much of it would be left un-combined?  =  40-36=  4g

10.0       (i) According to A  12g of C need  32 g of O2

According to B  24 g need .32g.   So  12 g need only  16g.

Therefore the correct equation is  B.

(ii) More oxygen has to be supplied.

11.0       3C2H2 + 3O2   --->  CO2 +CO + 4C +3H2O

3x 26  ………………………….4x12

1kg ……………………………=  48/ 78  =  615 g.

12.0       Equation:...........Zn  +      S    --->  Zn S

Formula mass …….. 65    +    32    =  97

6g of Zn need     32x6/ 65    of S=  3g

But supply is only 2g.

So we have to calculate the mass of ZNS using the mass of Sulphur.

32 g of S give…….97

2 g…………….   32x2/ 97=0.66g.