.For the given reaction, match column I with column II and mark the correct option from the codes given below.
Fe2O3 + xCO → yFe + xCO2
Column I Column II
(a) Oxidizing agent. (i) 2
(b) Reducing agent. (ii) 3
(c) x. (iii) Fe2O3
(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)
Fe2O3 + xCO → yFe + xCO2
Column I Column II
(a) Oxidizing agent. (i) 2
(b) Reducing agent. (ii) 3
(c) x. (iii) Fe2O3
(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)
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Answer:
Explanation: Column 2 B
Answer:
it's (D)
Explanation:
number of Fe atoms on LHS is 2
hence oh RHS is also 2 i.e, y=2
x should be equal to 3
oxidation state of Fe in Ferric oxide is +3 and on RHS it's 0(zero) - reduction
oxidising agent- which itself undergoes reduction
i.e, Fe2O3