Chemistry Paper 2 (Practical), WASSCE (PC 2ND), 2022

Question 2


(a)        How many atoms are there in 10.0 g of CaCO3?
[C = 12.0; O = 16.0; Ca = 40.0; Avogadro’s constant = 6.02 x 1023]


[5 marks]

(b)       

(i)         Write a balanced chemical equation to show the acidic nature of each of the
following compounds when reacted with sodium:
(I)        C2H2;
(II)       C2H5OH.

            (ii)        name the major  product in each of the reactions in 2(b)(i).
[6 marks]


(c)        Consider the following reaction scheme:

         

 

(i)         Draw the structures of compounds X and Y.
(ii)        What happens when bromine water is added to compound X and Y respectively.
(iii)       Explain the answer given in 2(c)(ii).

 

(d)        Magnesium reacts with aqueous solution of an acid to liberate hydrogen gas.
            Write the:

            (i)         half-reaction equations;
(ii)        overall reaction equation.
[3 marks]


(e)        (i)         State two differences between a mixture and a compound.
(ii)        Classify each of the following substances as an element, a mixture or a compound: quicklime; gold; soil.
An element:
A mixture:
A Compound:


Observation

Majority of the candidates responded to the question and their performance was average.

In part (a), few candidates calculated the number of atoms that are in 10.0 g of CaCO3.

In part (b), majority of the candidates could not write a balanced equation to show the acidic nature of C2H2 when reacted with sodium.

In part (c), majority of the candidates could not draw the structures of the compounds X and Y respectively.

In part (d), majority of the candidates could not write the half-reaction equation when magnesium reacts with aqueous solution of an acid to liberate hydrogen gas.

In part (e), majority of the candidates stated the differences between a mixture and a compound.
The expected answers include:

  (a)      Molar Mass of CaCO3 = 40 + 12 + 3 (16)     
= 100 g mol-1

                       :. 10.0 g CaCO3 = 10/100 

                                                 = 0.1 mol

            no. of atoms per molecule = 5
       ⇒ no of atoms in 10.0g CaCO3 = 5 x 0.1 x 6.02 x 1023
= 3.01 x 1023 atoms
                                                                                                                                   

(b)(i)           I.          C2H2 + 2Na → Na2C2(s) + H2(g) (2)
Or  2C2H2 + 2Na → 2C2HNa + H2
                   II.        2C2H5OH + 2Na → 2C2H5ONa(s) + H2(g)

(ii)      

I.          Sodium Carbide / Sodium ethynide   
II.        Sodium ethoxide

 

 

(ii)       Compound X decolourises bromine water but compound Y does not decolourise bromine water.

(iii)      Because compound X is an unsaturated compound and compound Y is a saturated compound

   

(d)(i)  Mg(s) →  Mg2+(aq)+ 2e-        (1)

           2H+(aq) + 2e- → H2(g)  (1)

      (ii)  Mg(s) + 2H+(aq) →  Mg2+(aq) + H2(g)(1)
                                                                       
(e)(i)


Mixture

Compound

-  Homogeneous or heterogeneous
-  Constituents are not chemically  
bound/cannot be represented by a
formula
-  Easily be separated/constituents retain 
their individual properties
-  Constituents can be added together in
any ratio by mass

-  Only homogeneous
-  Constituents are chemically   
bound/can be represented by a formula
-  Cannot easily be separated/loose their  
individual properties
-  Constituents are present in a fixed
ratio by mass

                                                                                                           
(ii)  quick lime – compound
Gold – element 
Soil – mixture