Atoms , Molecules & Stoichiometric ( Matric. 2011 & 2012 )

All prepared by alextan58@gmail.com

Question 1 ( Matriculation 2011 )

Calculate the molarity ( concentration ) of potassium chloride , KCl solution when

6.5g of KCl is dissolved in 500cm3 of water . [3m]

Answer :

Number of moles of KCl =

= 0.087 mol

Volume of water =

= 0.5 dm3

*1 dm3 = 1000 cm

3

Molarity / concentration =

mol dm

-3

= 0.174 mol dm-3

Atoms , Molecules & Stoichiometric ( Matric. 2011 & 2012 )

All prepared by alextan58@gmail.com

Question 2 ( Matriculation 2011 )

In a reaction , carbon dioxide , CO2 is produced when carbon monoxide , CO reacts

with oxygen according to the following reaction :

2CO (g) + O2 (g) → 2CO2 (g)

If 0.2 mole O2 is reacted with 0.36 mole CO , calculate the volume of CO2 produced

at the standard temperature and pressure . [4m]

Answer :

From the equation ,

1 mole of O2 = 2 mole of CO

0.2 mole of O2 = ( 0.2 × 2 ) mole of CO

= 0.4 mole of CO

Thus , carbon monoxide is the limiting reactant .

From the equation ,

2 mole of CO2 = 2 mole of CO

0.36 mole of CO2 = 0.36 mole of CO

Volume of CO2 produced = 0.36 × 22.4

= 8.06 dm3

Atoms , Molecules & Stoichiometric ( Matric. 2011 & 2012 )

All prepared by alextan58@gmail.com

Question 3 ( Matriculation 2011 )

The fertilizer urea , (NH2)2CO is prepared by reacting ammonia with carbon

dioxide :

2NH3 + CO2 → (NH2)2CO + H2O

If 12.5 g of NH3 is allowed to react with 25.0 g of CO2 . calculate

(a) the amount of (NH2)2CO ( in g ) produced [3m]

Answer :

Number of moles of NH3 =

mol

= 0.735 mol

Number of moles of CO2 =

mol

= 0.568 mol

From the equation ,

Number of moles of NH3 required = 2 × number of moles of CO2

= 2 × 0.568 mol

Since the number of moles of NH3 is less than that of required , thus limiting

reactant is NH3 .

From the equation ,

2 moles of NH3 = 1 mole (NH2)2CO

0.735 moles of NH3 =

mole of (NH2)2CO

= 0.368 mol

Mass of (NH2)2CO = 0.368 × [ 2(14.0) + 4(1.0) + 12.0 + 16.0 ]

= 22.1g

Atoms , Molecules & Stoichiometric ( Matric. 2011 & 2012 )

All prepared by alextan58@gmail.com

(b) the amount of excess reactant remained at the end of the reaction . [5m]

Answer :

From the equation ,

Number of moles of CO2 used =

× number of moles of NH3

=

× 0.735 mol

= 0.368 mol

Number of moles of CO2 remained = 0.568 – 0.368

= 0.20 mol

Mass of CO2 remained = 0.20 × [ 12.0 + 2(16.0) ]

= 8.80 g

Atoms , Molecules & Stoichiometric ( Matric. 2011 & 2012 )

All prepared by alextan58@gmail.com

Question 4 ( Matriculation 2012 )

The sub-atomic particles for the 4 particles are given in the following table . The

particles maybe an atom or ion .

Particle Protons Neutrons Electrons

P 13 14 10

Q 8 9 8

R 13 14 13

S 8 8 8

(a) Select 2 particles which are atom and ion of the same element . Explain your

answer . [2m]

Answer :

P and R

They have same number of protons but different in number of electrons

(b) Select 2 particles which are isotopes of the same element . Explain your answer .

[2m]

Answer :

Q & S

They have same number of protons but different in number of neutrons .