Molybdenum metal, used in special types of steel a always can be manufactured by

Question

Molybdenum metal, used in special types of steel a always can be manufactured by the reaction of its oxide with hydrogen gas. MoO_3(s) + 3H_2(g) rightarrow Mo(s0 + 3H_2O(s) What is the oxidation # of Mo in MoO_3(s)? b) What is the oxidation # of O in MoO_3(s)? c)What is the oxidation # of H in H_2(g)? d) What is the oxidation # of Mo in Mo(s)? e) What is the oxidation # of H in H_2O? f) What is the oxidation # of O in H_2O(e)? What element is oxidized? What is the oxidizing agent h) What element is reduced? What is the reducing agent?

Explanation / Answer

a) Oxidation state of Mo be x

x + 2(-3) = 0

x = 6

Hence oxidation state of Mo will be equal to +6 in MoO3

b) Oxidation state of O will be -2 in MoO3

c) Oxidation state of H in H2 is equal to 0

d) Oxidation state of Mo in Mo(s) will be equal to 0

e) Oxidation state of H in H2O be x

2x - 2 = 0

x = 1

Hence oxidation state of H in H2O will be equal to 1

f) Oxidation state of O in H2O is equal to -2

g) Hydrogen is oxidized, MoO3 is oxidizing agent

h) Mo is reduced, H2 is the reducing agent

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