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9.Using Equations.

posted Mar 28, 2016, 11:08 PM by Upali Salpadoru   [ updated Dec 23, 2016, 11:07 AM ]

 Making use of Moles.

Ox2.jpg






Fig.1 Lab preparation of Oxygen.

Some manganese dioxide speeds up the reaction.

In finding the mass of reactants and products it is very convenient to work in moles.

As the equations give the number of participating molecules and the molar mass is equal to 1 mole the task becomes very simple.

Let’s take an example;      

 

Reactants

Products

Word Equation

Potassium chlorate

Potassium chloride

Oxygen

Equation

2KClO3

2KCl                                     

3 O2

No. of molecules      

2

2(+ions) +  2 (_ ions)

3

No. of moles.

2

2(+ions) +  2 (_ ions)

3



Example 1.

Find mass of oxygen from 1 mol of potassium chlorate.

                       2mol of potassium chlorate give  ….3 mol of oxygen.

                        1 mol  ……………………………..3/2  mol of oxygen.

Mass of 1 mol of oxygen = 32g

Mass of 3/2 mol………..= 3/2 x 32 g

                                       =  48g.

Example 2

Oxygen and hydrogen can be made to combine by an electric spark’ Find the mass of water formed when the mixture had  2g of Hydrogen and 2 g. Of oxygen.

Converting the given masses to moles:

 Hydrogen =2/2 =1 mol.     Oxygen   2/32 = 0.06 mol.

As 1 mol of Hydrogen would require 0.05 mol of Oxygen, the given mass of Oxygen would be able to react.

Equation:     Hydrogen  + Oxygen  ----->  water

                      2 H2          +   O2     ------>   2H2O                      

Moles             2mol.     +  1mol     ------>    2mol

                    1mol of  oxygen makes…….2 mol of water.

                    2/32…... (0.06)  ……….. =  2x0.06   =   0.12 mol

            Mass of 1 mol of water = 18g

               Mass of 0.12 mol of water =  18x0.12 =2.16 g

                    Remember:  Converting mass to moles.

Moles =    Mass /  Molar mass

                                          Converting moles to mass                           

Mass  =  Moles x  Molar mass.

Practice.jpg

[  H – 1,  Cl – 35.5 ,  C – 12,  Ca – 40, Pb-  207   S -32,   N – 14, ]

1. 0 Convert the given masses to mol

mass

36.5g of  HCl

440g of CO2

1.8 g  of H20

1kg of CaCO3

mol

 

 

 

 

 2.0 Convert these mol to mass.

mol

2 of NaOH

2 of Pb(NO3)2 

0.1 of H2SO4

100  of S

Mass

 

 

 

 


                     

3.0 Give the number of mol that will result during these reactions.

 

Reacting

Resulting

i)           Equation

        H2      +           Cl2 

           2 HCl

Reacting and resulting mol ratio

……………+ ………..

 ……………….


ii)          Equation

    C           +       O2

    CO2

Reacting and resulting mol ratio

………..   +   …………

 ………………


iii)         Equation

2NaOH     +   H2SO4

        Na2SO4        +     2 H2O

Reacting and resulting mol ratio

 

 

 

iv)        Equation

Pb(NO3)2  +  2 NaCl  

  PbCl2        +            2NaNO3

Reacting and resulting mol ratio

 

 

 

v)            Equation

2Al         +      6HCl

    2AlCl3         +  3H2

Reacting and resulting mol ratio

 

 

4,0 

i)  If 3 mol of Hydrogen and 1mol of Chlorine were allowed to react , find the number of

mol of HCl produced . ………………

ii) If 6 g of Carbon was burnt in air, find the mass of CO2 formed…………………….

iii) If 40 g of NaOH reacts with H2SO4 find the maximum mass of Na2SO4 that can be recovered.

Iv) Using this equation,  find the mass of CaO formed from 1kg of lime stone.(CaCO3.

       CaCO3       

CaO      +  CO2

v)Using this equation, find the mass of the precipitate PbCl2 formed if 2 mol of Pb(NO3)2  reacts with 1 mol of NaCl.

Pb(NO3)2  +  2 NaCl  

PbCl2        +            2NaNO3

331          +   2 x 58.5

   278           +        2 x 85

 5.0

i)   A solution of salt containing 1 mol dissolved n a litre was evaporated . Find the maximum number of mol. that will form the residue.

ii)  If 100 ml. of an acid solution had 0.5 mol in it , find the number of mol in a litre of that solution.

 
 

ANSWERS

1.0

2.0

3.0

4.0

5.0

i.)1

ii)10

iii)0.1

iv) 10

i)80g

ii)662g

iii)9.8

iv)3.2kg

i)1:1:2

ii)1:1:1

iii)2:1:1:2

iv)1:2:1:2

v2:6:2:3

i) 2

ii)22g

iii)71g

iv)560g

v)139g

i) 1 mol

ii) 5 mol


 






Q.1.0

 If you are given 6.g of Hydrogen and 71.g. of chlorine for the reaction given, find the following.  

Equation:     H2   +Cl2  - ----->       2 HCl

1.1  Molar ratio of hydrogen and oxygen required.

1.2  Molar  ratio of hydrogen and oxygen given.

1.3  Maximum mass of Hydrogen chloride formed.

              (5x3=15 )

Q. 2.0.

Carbon burns to form carbon dioxide.

Equation:      C   +  O2  ----->   CO2

2.1 Obtain the molar ratios of reactants and the product.

2.2 Convert 6g of Carbon to moles.

2.3 Find the number of moles of CO2 you may get from 6g of carbon.

2.4 Convert the above answer to mass of CO2.

2.5 Assuming that 1 mol  of a gas can occupy 25 litres, under the room temperature and pressure find the volume of CO2 formed.

                                           (4x5=20 )

Q.3.0

A solution X contains 33.1g. of lead nitrate dissolved in 500 ml.

A solution Y contains 5.85g of sodium chloride dissolved in 1000 ml.

They react forming a precipitate of lead chloride.

Equation:   Pb(NO3)2 + 2  NaCl  ------>  PbCl2  + 2  NaNO3


3.1 How many moles of lead nitrate are there in the X solution?

3.2 How many moles of sodium chloride are there in the Y solution?

3.3  If X and Y are mixed the reacting Molar ratios.  Lead nitrate : sodium chloride.

3.4 How many moles of lead chloride will precipitate during the reaction?

3.5 How many moles of the following will remain in the solution?

           Lead nitrate…….  Sodium nitrate……..sodium chloride……….

                                                                                    (3x5=15)

Q. 4.0

By heating corals it is possible to obtain carbon dioxide used in aerated drinks and quicklime used to make cement.

Equation:- CaCO3  ----->   CaO  +CO2

Find the following:

4.1 Reacting mole ratios:  CaCO3  : CaO  : CO2  =  ……. : .......... : ...........

4.2 The required volume of carbon dioxide at the required temperature and pressure is 36 litres . If 1 mol of the gas occupies 12 litres,under given conditions, find the required quantity in moles.

4.3  How many moles of CaCO3   would be necessary for this?

4.4  What would be the mass of  CaCO3  required?

4.5  What would be the mass of CaO  formed?

Q. 5.0

A student suspended 15 g of cleaned iron nails in a blue coloured solution of copper sulphate. The next day he observed the solution to be colourless and grey coloured iron had a coating of copper.  When the iron was cleaned and weighed it was found to be 9.4 g.

Equation   ..Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq)

Molar ratios  1……...1……………...1……...1

Find the following:

5.1 The mass of iron that has reacted.

5.2  Number of moles of iron reacted.

5.3 Number of copper  moles produced.

5.4 Mass of copper formed.

5.5 Number of copper sulphate moles reacted?

                                            (3x5=15)

Q. 6.0

Aqueous Sodium hydroxide reacts with aqueous copper sulphate forming a precipitate of copper hydroxide. A litre of a solution of sodium hydroxide contains 3 moles.

  Equation:-     CuSO4 + 2 NaOH ----->  Cu(OH)2 + Na2SO4

Find the following:-

6.1  Number of moles of NaOH in 100ml of the solution.

6.2  Number of moles of copper sulphate required to remove all the OH- ions in 100ml of NaOH.

6.3 Number of moles of copper hydroxide that can form with this?

6.4 Mass of copper hydroxide formed.?

6.5 Mass of metallic copper that can be obtained from this precipitate.?

                                 (3x5=15)

7.0

Excess of sulphuric acid was added to 500 ml of  sodium hydroxide solution. When the mixture was evaporated the mass of the residue was 1.4g.

7.1 Find the moles of the Na2SO4formed.

7.2 Determine the strength of the alkali. (Moles per liter)

Equation:-   2NaOH  + H2SO4  ---->Na2SO4  +  2H2O

                                      (2x10=20)
For answers click Answer page.
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