Question

$wT{l}_2{O}_3\left(s\right)+xN{H}_{2}OH\left(aq\right)\to yTlOH\left(s\right)+z{N}_2\left(g\right)$ where $w,x,y,z$ are the stoichiometric coefficients of $T{l}_2{O}_3,N{H}_{2}OH,TlOH,N_2$ respectively.
Choose the correct statements:

• A. $\frac{y}{z}=2$
• B. $\frac{w}{x}=0.25$
• C. $\frac{w}{x}=0.5$
• D. $\frac{y}{z}=1$

Answer 1

Answer: (B), (D)

$\frac{y}{z}=1$

$T{l}_2{O}_3\left(s\right)+N{H}_{2}OH\left(aq\right)\to TlOH\left(s\right)+N_2\left(g\right)$

Oxidation state of $Tl$ in $T{l}_2{O}_3=+3$
Oxidation state of $Tl$ in $TlOH=+1$
Oxidation state of $N$ in $N{H}_{2}OH=-1$
Oxidation state of $N$ in $N_2=0$

$N$ is undergoing oxidation.
$Tl$ is undergoing reduction.

Finding $n_f:$
$n_f$ of $T{l}_2{O}_3=4$
$n_f$ of $N{H}_{2}OH=1$

Cross multiplying these with $n_f$ of each other.
we get,
$T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to TlOH\left(s\right)+N_2\left(g\right)$

Balancing the elements other than oxygen and hydrogen on both sides,
$T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+2N_2\left(g\right)$

Adding the $H_{2}O$ to balance the oxygen,

$T{l}_2{O}_3\left(s\right)+4N{H}_{2}OH\left(aq\right)\to 2TlOH\left(s\right)+2N_2\left(g\right)+5H_{2}O$

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.
$\frac{w}{x}=\frac{1}{4}=0.25\frac{y}{z}=\frac{2}{2}=1$