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Question

wTl2O3(s)+xNH2OH(aq)yTlOH(s)+zN2(g) where w,x,y,z are the stoichiometric coefficients of Tl2O3, NH2OH, TlOH, N2 respectively.
Choose the correct statements:

A
yz=2
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B
wx=0.25
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C
wx=0.5
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D
yz=1
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Solution

The correct options are
B wx=0.25
D yz=1
Tl2O3(s)+NH2OH(aq)TlOH(s)+N2(g)

Oxidation state of Tl in Tl2O3=+3
Oxidation state of Tl in TlOH=+1
Oxidation state of N in NH2OH=1
Oxidation state of N in N2=0

N is undergoing oxidation.
Tl is undergoing reduction.

Finding nf:
nf of Tl2O3=4
nf of NH2OH=1

Cross multiplying these with nf of each other.
we get,
Tl2O3(s)+4NH2OH(aq)TlOH(s)+N2(g)

Balancing the elements other than oxygen and hydrogen on both sides,
Tl2O3(s)+4NH2OH(aq)2TlOH(s)+2N2(g)

Adding the H2O to balance the oxygen,

Tl2O3(s)+4NH2OH(aq)2TlOH(s)+2N2(g)+5H2O

Hydrogen is already balanced.
Charges on both sides is also balanced.
So, it is the final balanced equation.

So, the value of w,x,y,z is 1,4,2,2 respectively.
wx=14=0.25yz=22=1

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