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Karnataka Board PUCPUC Science Class 11

A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over heated element (1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent

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Question

A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over heated element (1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric oxide to iron. Identify the element and write formulas of its monoxide and dioxide. Write chemical equations for the formation of producer gas and reduction of ferric oxide with the monoxide.

Long Answer
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Solution

\[\ce{2C (s) + O2 + 4N2 (g) ->[123 k] 2CO (g) + 4N2 (g)}\]

\[\ce{Fe2O3 (S) + 3CO (g) ->[heat] 2Fe (S) + 3CO2}\]

C = tetravalent carbon 

CO = carbon monoxide 

Fe2O3 = ferric oxide 

CO2 = carbon dioxide

Tetravalent elements i.e. carbon combines with oxygen to produce carbon monoxide. The reaction occurs at high temperatures and in the presence of nitrogen gas, which acts as a producer gas only. It is not consumed in the reaction. The carbon monoxide formed acts as a reducing agent for ferric oxide, reduces the oxidation state of iron from +3 to zero, and oxidizes itself to form carbon dioxide.

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Chapter 11: The p-block Elements - Multiple Choice Questions (Type - I) [Page 141]

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NCERT Exemplar Chemistry Exemplar [English] Class 11
Chapter 11 The p-block Elements
Multiple Choice Questions (Type - I) | Q 50 | Page 141

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