Mole Concept Solutions

Go to the Tutorials on Mole Concept

QUESTION 1
How many moles of limestone, CaCO3 will be required to produce 5.6 g of quicklime, CaO?

                        CaCO3 (s) → CaO (s) + CO2 (g)

[Ca = 40, C = 12, O = 16]

A. 0.20
B. 0.10
C. 1.12
D. 0.56

SOLUTION
From the question, information is only provided about calcium oxide and we are to use that to get the moles of limestone that reacted. This means that the major players are CaCO3 and CaO, carbon (IV) oxide is a detractor/spectator.

Hence, from the equation of reaction:

                                     1 mol CaCO3 ≡ 1 mol CaO …………. (I)

                            number of moles, n = reacting mass, m/molar mass, M
                      
                                                nCaO = mCaO/MCaO

                                                         = 5.6/(40 + 16)
                                                         = 5.6 g/56 g mol-1
                                                         = 0.1 mol

From equation (i),                   nCaCO3 = nCaO

therefore,                               nCaCO3 = 0.10 mol    [B]

Alternatively, we can use the mole-mass relationship to solve the problem:

                               1 mol CaCO3 ≡ 1 mol CaO (mole ratio) …………. (I)

                               1 mol CaCO3 ≡ 56 g CaO (mole-molar mass relationship)
                              
                               x mol CaCO3 ≡ 5.6 g mol-1 CaO (mole-reacting mass)

Solving for x;                             x = 5.6 g/5.6 g mol-1
                                                   
                                                   = 0.10 mol     [B]

QUESTION 2
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

What volume of gas is evolved at s.t.p if 2 g of calcium trioxocarbonate (IV) is added to a solution of hydrochloric acid?

[Ca = 40, C = 12, O =16, Cl = 35.5, H = 1, Molar volume of a gas at s.t.p = 22.4 dm3]

A. 112 cm3
B. 224 cm3
C. 448 cm3
D. 2240 cm3

SOLUTION
The question provides information on the mass of calcium trioxocarbonate (IV), and we are meant to calculate the volume of gas (carbon (IV) oxide) that was evolved. So, the major players of the reaction are CaCO3 and CO2, while HCl, CaCl2 and H2O are the detractors/spectators.

From the equation of reaction:

                                             1 mol CO2 ≡ 1 mol CaCO3 …………. (I)

                                 number of moles, n = reacting mass, m/molar mass, M

                                                 nCaCO3 = m CaCO3/M CaCO3

                                                              = 2/[40 + 12 + (16 x 3)]

                                                              = 2 g/100 g mol-1

                                                              = 0.02 mol

From equation (I),                           nCO2 = nCaCO3

therefore,                                        nCO2 = 0.02 mol

but,                             number of moles, n = volume, v/molar volume, Vm

                                                       vCO2 = nCO2 x VmCO2

                                                                = 0.02 mol x 22.4 dm3 mol-1
           
                                                                = 0.448 dm3 x 1000 cm3 dm-3

                                                                = 448 cm3   [C]

Note that whenever you use the value of the molar volume of a gas at s.t.p as 22.4 dm3, the unit of the calculated volume will be in dm3.

In this question, the required unit of the calculated volume is cm3, which is why it is to converted from dm3 to cm3 by using the conversion of 1000 cm3 = 1 dm3

Alternatively, we can use the mass-volume relationship:

                          1 mol CaCO3 ≡ 1 mol CO2 (mole ratio) …………. (I)

                          100 g CaCO3 ≡ 22.4 dm3 CaCO3 (molar mass-molar volume)

                              2 g CaCO3 ≡ x dm3 CO2 (mass-volume)

Solving for x;

                                           x = 2 g x 22.4 dm3/100 g

                                              = 0.448 dm3 x 1000 cm3 dm-3

                                              = 448 cm3   [C]


QUESTION 3
If a solution contains 4.9 g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it. [Cu = 64, O = 16, S = 32, H = 1]

A. 0.8 g
B. 4.0 g
C. 40.0 g
D. 80.0 g

SOLUTION
No equation of reaction is given in this question. So, the first thing is to write the balanced equation of reaction.

                            H2SO4 (aq) + CuO (s) → CuSO4 (aq) + H2O (l)

Then, separate the players from the spectators. H2SO4 and CuO are the players because we are asked to use the mass of H2SO4 to determine the mass of CuO that reacted. Nothing was said about CuSO4 and H2O, which means they are the spectators.

So, from the equation of reaction:

                                  1 mol H2SO4 ≡ 1 mol CuO …………. (I)

                         number of moles, n = reacting mass, m/molar mass, M

                                          nH2SO4 = m H2SO4/M H2SO4

                                                       = 4.9/[(2 x 1) + 32 + (16 x 4)]

                                                       = 4.9 g/98 g mol-1

                                                       = 0.05 mol

From equation (I),                    nCuO = nH2SO4

therefore,                                nCuO = 0.05 mol

but,
                                              mCuO = nCuO x MCuO

                                                        = 0.05 x (64 + 16)

                                                        = 0.05 mol x 80 g mol-1

                                                        = 4.0 g   [B]

Alternatively, we can use the mass-mass relationship:

                                     1 mol H2SO4 ≡ 1 mol CuO (mole ratio) …………. (I)

                                       98 g H2SO4 ≡ 80 g CuO (molar mass-molar mass)

                                       4.9 g H2SO4 ≡ x g CuO (mass-mass)
Solving for x;

                                                        x = 4.9 g x 80 g mol-1/98 g mol-1

                                                           = 4.0 g   [B]