\[F{{e}_{2}}{{O}_{3}}+xCO\to yFe+xC{{O}_{2}}\] |
Column I | Column II |
(a) Oxidizing agent | (i) 2 |
(b) Reducing agent | (ii) 3 |
(c)\[x\] | (iii)\[F{{e}_{2}}{{O}_{3}}\] |
(d) y | (iv) CO |
A) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
B) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
C) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
D) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Correct Answer: C
Solution :
Balanced chemical equation is \[F{{e}_{2}}{{O}_{3}}+3CO\xrightarrow{{}}2Fe+3C{{O}_{2}}\] Thus, \[x\] is 3 and y is 2. |
In this reaction, \[F{{e}_{2}}{{O}_{3}}\]loses oxygen so, it gets reduced and acts as an oxidizing agent while CO gains oxygen so, it gets oxidized and acts as a reducing agent. |
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