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In the given reaction $PbS\xrightarrow[{Roasting}]{{air/\Delta }}X$, $X + PbS \to Pb + S{O_2}$
X is:
A. $PbO$
B. $Pb{O_2}$
C. $PbO$ and $PbS{O_4}$
D. $Pb{O_2}$ and $PbO$

Last updated date: 20th Jun 2024
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Hint: The process of heating of sulphide ore in the presence of a sufficient amount of oxygen at very high temperature is known as roasting. During the process of roasting the sulphide ores are converted into their corresponding oxide with the release of sulphur dioxide.

Complete step by step answer:
When lead sulphide is roasted in the presence of air at high temperature there is the formation of lead oxide take place and this lead oxide reacts with lead sulphide to form lead and sulphur dioxide. Thus the above reaction is written as:

$PbS\xrightarrow[{Roasting}]{{air/\Delta }}PbO$

$PbO + PbS \to Pb + S{O_2}$

So in the above question X is lead oxide.

So, the correct answer is Option A.

Additional information:
lead oxide (PbO) is a monoxide. It exists in two polymorphs such as litharge which have a tetragonal crystal structure and massicot exist in the orthorhombic crystal structure. Lead monoxide is used for making the glass because it can decrease the viscosity of glass and increase the electrical resistivity of glass. It also increases the ability of a glass to absorb X-ray radiation. It is also used in the vulcanization of rubber and production of certain pigments and paints. It is also a key component of automotive lead-acid batteries.

Note: Lead monoxides are amphoteric which means that it can react both acid as well as a base also. When it reacts with acid it forms the salt of $P{b^{2 + }}$ion via the intermediacy of oxo cluster whereas with strong base lead monoxide dissolve to form plumbite which is also called plumbate(ii) salts.