What is the volume of chlorine gas produced at s.t.p., in \text{dm}^3, when 8.7\text{ g} o… — JAMB UTME

What is the volume of chlorine gas produced at s.t.p., in $\text{dm}^3$ , when $8.7\text{ g}$ of manganese(IV) oxide ( $MnO_2$ ) reacts completely with excess concentrated hydrochloric acid ( $HCl$ )? [Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{mol}^{-1}$ , $Mn = 55$ , $O = 16$ ]

Answer: 2.24 dm³

Answer

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

2.24\text{ dm}^3

In the laboratory preparation of chlorine, manganese(IV) oxide acts as an oxidizing agent according to

MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O

. Since

8.7\text{ g}

MnO_2

represents

0.1\text{ mol}

, exactly

0.1\text{ mol}

Cl_2

gas is produced. At s.t.p.,

0.1\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 2.24\text{ dm}^3

Step-by-Step Solution

Write the balanced equation for the preparation of chlorine gas from manganese(IV) oxide and concentrated hydrochloric acid.

MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O

Establishes the stoichiometric relationship between

MnO_2

and

Cl_2

Calculate the molar mass of manganese(IV) oxide (

MnO_2

87\text{ g mol}^{-1}

Required to convert the given mass of

MnO_2

into moles.

Calculate the moles of

MnO_2

supplied.

\frac{8.7\text{ g}}{87\text{ g mol}^{-1}} = 0.1\text{ mol}

Determines the exact amount of reactant taking part in the reaction.

Multiply moles of

Cl_2

by the molar volume of a gas at s.t.p.

0.1\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

One mole of any ideal gas occupies

22.4\text{ dm}^3

at s.t.p.

Key Concept

Laboratory preparation and stoichiometry of chlorine gas evolution.

Estimated Time:1m 30s

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