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9966 [12]
3 years ago
5

In the reaction Mg(s) + 2H2O(l) --> Mg(OH)2(s) + H2(g), what is the oxidation state of each H in H2(g)? A. +2 and –2 B. –1 an

d +1 C. +1 D. 0 no idea how to do this. I know its not D because that's for magnesium
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
6 0
The oxidation state of the atoms of any molecule in the element state equals zero. H2 is is the element state, therefore the oxidation state of the whole molecule equals zero and the oxidation states of the atoms also equals zero. Answer letter D is the correct answer. 
Magnesium has an oxidation state = zero before it undergoes the reaction and is oxidized. The oxidations state of the magnesium in Mg(OH)2 is +2.
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(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X
 
           = 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M


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