Potassium tetraoxomanganate(VII), \text{KMnO}_4, reacts with excess concentrated hydrochlo… — JAMB UTME

Potassium tetraoxomanganate(VII),

\text{KMnO}_4

, reacts with excess concentrated hydrochloric acid according to the redox equation:

2\text{KMnO}_4(s) + 16\text{HCl}(aq) \rightarrow 2\text{KCl}(aq) + 2\text{MnCl}_2(aq) + 8\text{H}_2\text{O}(l) + 5\text{Cl}_2(g)

Calculate the volume of chlorine gas (in

\text{dm}^3

) produced at s.t.p. when

15.8\text{ g}

\text{KMnO}_4

reacts completely. [Molar mass of

\text{KMnO}_4 = 158\text{ g mol}^{-1}

, molar volume of gas at s.t.p. =

22.4\text{ dm}^3\text{mol}^{-1}

]

Cevap: 5.6 dm^3

The volume of chlorine gas liberated at s.t.p. is

5.6\text{ dm}^3

Converting

15.8\text{ g}

\text{KMnO}_4

gives

0.10\text{ mol}

. According to the balanced equation,

2\text{ mol}

\text{KMnO}_4

yields

5\text{ mol}

\text{Cl}_2

, giving

0.25\text{ mol}

\text{Cl}_2

. Multiplying

0.25\text{ mol}

by the molar volume at s.t.p. (

22.4\text{ dm}^3\text{mol}^{-1}

) yields

5.6\text{ dm}^3

Calculate the moles of potassium tetraoxomanganate(VII),

\text{KMnO}_4

\text{Moles of KMnO}_4 = \frac{15.8\text{ g}}{158\text{ g mol}^{-1}} = 0.10\text{ mol}

Converting the given mass of reactant to moles allows stoichiometric comparison.

Use the balanced redox equation to determine the mole ratio between

\text{KMnO}_4

and

\text{Cl}_2

\text{Moles of Cl}_2 = 0.10\text{ mol} \times \frac{5}{2} = 0.25\text{ mol}

The equation shows that

2\text{ moles}

\text{KMnO}_4

produce

5\text{ moles}

\text{Cl}_2

gas.

Calculate the volume of

\text{Cl}_2

gas produced at s.t.p.

\text{Volume} = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 5.6\text{ dm}^3

At s.t.p.,

1\text{ mole}

of any gas occupies

22.4\text{ dm}^3

Stoichiometric calculations in the laboratory preparation of chlorine from oxidation of hydrochloric acid

Soru