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Q.
The ratio of $\ce{Fe^{3+}}$ and $\ce{Fe^{2+}}$ in $\ce{Fe_{0.9} S_{1.0}}$ is
Redox Reactions
Solution:
$Fe ^{3+}= x$
$Fe ^{2+}=0.93- x$
Using charge balance
$Fe _{0.9} S _{1.0}=$ Charge on neutral compound $=0$
$\underbrace{3 \times x 2 \times(0.93-x)}+2 \times 1=0$
Charge on $Fe _{0.9}$
$\left( Fe ^{2+} \& Fe ^{3+}\right)$
$\Rightarrow 3 x +2(0.9- x )=2 \times 2$
$3 x+1.8-2 x=2$
$x =2-1.8$
$x =0.2$
Thus the no. of $Fe ^{+2}$ ion $=0.93- x$ $=0.93-0.2$ $=0.73$
Ratio $\frac{ Fe ^{3+}}{ Fe ^{2+}}=\frac{0.2}{0.73}$ $=0.28$