STOICHIOMETRIC RELATIONSHIPS, Therefore, mole molecules, SOLUTION, Number of moles

 

STOICHIOMETRIC RELATIONSHIPS :

 

            Calculation by using moles and balanced chemical equations reacting or produced masses, volume of gases, volume and concentration of solution are called stoichiometric relationships.

 

PROBLEM:

 

            When ICl₃ dissolved in excess of aqueous Kl, iodine is liberated.

(a)              Predict a chemical equation for the reaction between ICl₃ and Kl.

(b)              What volume of 1.00 mole dm^-3 sodium thiosulphate is required to react with all the iodine liberated when 1.00g of ICl₃ reacts with an excess of Kl.

SOLUTION:

            (a)       ICl3 + 3Kl                               3KCl + 2l2

            (b)       mole molecules of ICl3 = 1 / 233.5

                                                                = 4.283 x 10^-3 moles.

From the equation, mole ratio

                        ICl₃   :  l₃

                           1     :   2       

Therefore, mole molecules of I2 = 2 (4.283 x 10-3)

                                                        = 8.565 x 10-3 moles.

            2Na₂S₂O₃ + I₂                         Na₂S₄O₆+2Nal

                        2          :           1

Therefore, moles of Na2S2O3 required = 2(8.565 x 10^-3)

        = 0.0171 moles.

            V = n x 1000 / M

            V = 0.0171 x 1000 / 1.00

                = 17.1 cm³

 

PROBLEM:

            Sulpher and chlorine react together to form S₂Cl₂. When 1.0g of this S₂Cl₂ react with water, 0.36g of a yellow ppt. (sulpher) was formed together with a solution containing a mixture of sulpher, H₂SO₄ and HCl.

 

(a)       By using the given data suggest a balanced chemical equation for the reaction between S₂Cl₂ and H₂O.

(b)       What volume of 1.00 mole dm^-3 NaOH is required to neutralize the final solution?

 

SOLUTION:

(a)              Mole ratio between S₂Cl₂ and sulpher (S) is,

S₂Cl₂         :      S

1 / 135     :     0.36 / 32

0.00741    :     0.01125

            1    :     1.5

            2    :     3

            2S₂Cl₂ + ₃H₂O                         3S + H₂SO₃ + 4HCl

                                 ^{(l)                  (l)                                                            (s)                   (aq)                (aq)}

            (b)       H₂SO₃ + 4HCl + 6NaOH                                 Na₂SO₃ + 4NaCl + 5H₂O

 

                        Therefore, mole ratio

                                                                        S₂Cl₂    :     NaOH

                                                                             2     :        6

                                                                             1     :        3

                        Moles of NaOH = 3(0.00741)

                                                  = 0.02223 moles.

                                                V = n x 1000 / M

                                                  = 0.02223 x 1000 / 1

                                                  = 22.23 cm³ of NaOH

PROBLEM:

            A nail of mass 1.40g was dissolved in a excess of dilute H₂SO₄ to form 100 cm³ of solution. A 10 cm³ sample  of t his solution required 4.0 x 10^-4 moles of mangnate (vii) for complete oxidation. It is assume that iron present in nail was completely converted into Fe²⁺ while dissolving in dilute sulphuric acid. Calculate the number of moles of Fe²⁺ produce by the nail and also calculate the %age by mass of iron in the nail.

 

SOLUTION:

            MnO₄^1- + 8H¹⁺ + 5Fe²⁺                                  Mn²⁺ + 4H₂O + 5Fe³⁺

 

Mole ratio between,

                        MnO₄^1-     :     Fe²⁺

                1           :      5         

Therefore,

Number of moles of Fe²⁺ in 10 cm³ of solution = 5 (4.0 x 10^-4)

                                                                               = 2 x 10^-3 moles 

Number of moles of Fe²⁺ in 100 cm³ of solution = 10 (2 x 10^-3)

                                                                                  =2.0 x 10^-2 moles.

Mass of Fe²⁺ (iron) = moles x Ar

                        = 2.0 x 10^-2 x 56

                        = 1.12g

%age by mass of iron in the nail = 1.12 / 1.40 x 100

                                                        = 80%