Manganese (IV) oxide, lead (IV) oxide and red lead [$P{b_3}{O_4}$] reacts with conc. HCl liberating chlorine. What is the common property being shown by these metal oxide.
Answer
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Hint: The metal oxide reacts with hydrochloric acid to give metal chloride, water and chlorine gas. In this reaction oxidation and reduction is both taking place, where hydrochloric acid is being oxidized to chlorine gas.
Complete step by step answer:
It is given that manganese (IV) oxide, lead (IV) oxide and red lead [$P{b_3}{O_4}$] reacts with concentration hydrochloric acid to give chlorine gas.
The reaction of Manganese (IV) oxide $Mn{O_2}$ with hydrochloric acid HCl is shown below.
$Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}$
In this reaction, one mole of manganese (IV) oxide reacts with four mole of hydrochloric acid to give one mole of manganese chloride, two mole of water and one mole of chlorine gas.
In this reaction, we can see that manganese (IV) oxide acts as an oxidizing agent which oxidizes hydrochloric acid to form chlorine gas.
The reaction of lead (IV) oxide with hydrochloric acid is shown below.
$Pb{O_2} + 4HCl \to PbC{l_2} + 2{H_2}O + C{l_2}$
In this reaction, one mole of lead (IV) oxide reacts with four mole of hydrochloric acid to give one mole of lead chloride, two mole of water and one mole of chlorine gas.
In this reaction, we can see that lead (IV) oxide acts as an oxidizing agent which oxidizes hydrochloric acid to form chlorine gas.
The reaction of red lead $P{b_3}{O_4}$ with hydrochloric acid is shown below.
$P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + 4{H_2}O + C{l_2}$
In this reaction, one mole of red lead reacts with eight mole of hydrochloric acid to give three mole of lead chloride, four mole of water and one mole of chlorine gas.
In this reaction, we can see that red lead acts as an oxidizing agent which oxidizes hydrochloric acid to form chlorine gas.
Therefore, the common property shown by the metal oxide is that they act as an oxidizing agent.
Note: In this reaction hydrochloric acid acts as a reducing agent which reduces the metal oxide to metal chloride by gaining electrons. The change of metal oxide to metal chloride is a reduction reaction.
Complete step by step answer:
It is given that manganese (IV) oxide, lead (IV) oxide and red lead [$P{b_3}{O_4}$] reacts with concentration hydrochloric acid to give chlorine gas.
The reaction of Manganese (IV) oxide $Mn{O_2}$ with hydrochloric acid HCl is shown below.
$Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}$
In this reaction, one mole of manganese (IV) oxide reacts with four mole of hydrochloric acid to give one mole of manganese chloride, two mole of water and one mole of chlorine gas.
In this reaction, we can see that manganese (IV) oxide acts as an oxidizing agent which oxidizes hydrochloric acid to form chlorine gas.
The reaction of lead (IV) oxide with hydrochloric acid is shown below.
$Pb{O_2} + 4HCl \to PbC{l_2} + 2{H_2}O + C{l_2}$
In this reaction, one mole of lead (IV) oxide reacts with four mole of hydrochloric acid to give one mole of lead chloride, two mole of water and one mole of chlorine gas.
In this reaction, we can see that lead (IV) oxide acts as an oxidizing agent which oxidizes hydrochloric acid to form chlorine gas.
The reaction of red lead $P{b_3}{O_4}$ with hydrochloric acid is shown below.
$P{b_3}{O_4} + 8HCl \to 3PbC{l_2} + 4{H_2}O + C{l_2}$
In this reaction, one mole of red lead reacts with eight mole of hydrochloric acid to give three mole of lead chloride, four mole of water and one mole of chlorine gas.
In this reaction, we can see that red lead acts as an oxidizing agent which oxidizes hydrochloric acid to form chlorine gas.
Therefore, the common property shown by the metal oxide is that they act as an oxidizing agent.
Note: In this reaction hydrochloric acid acts as a reducing agent which reduces the metal oxide to metal chloride by gaining electrons. The change of metal oxide to metal chloride is a reduction reaction.
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