What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide

Let the maximum concentration of each solution be $x \mathrm{~mol} / \mathrm{L} .$ After mixing, the volume of the concentrations of each solution will be reduced to half i.e., $\frac{x}{2}$.

$\therefore\left[\mathrm{FeSO}_{4}\right]=\left[\mathrm{Na}_{2} \mathrm{~S}\right]=\frac{x}{2} \mathrm{M}$

Then, $\left[\mathrm{Fe}^{2+}\right]=\left[\mathrm{FeSO}_{4}\right]=\frac{x}{2} \mathrm{M}$

Also, $\left[\mathrm{S}^{2-}\right]=\left[\mathrm{Na}_{2} \mathrm{~S}\right]=\frac{x}{2} \mathrm{M}$

$\mathrm{FeS}_{(s)} \longleftrightarrow \mathrm{Fe}^{2+}{ }_{(a q)}+\mathrm{S}^{2-}{ }_{(a)}$

$K_{s p}=\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{S}^{2-}\right]$

$6.3 \times 10^{-18}=\left(\frac{x}{2}\right)\left(\frac{x}{2}\right)$

$\frac{x^{2}}{4}=6.3 \times 10^{-18}$

$\Rightarrow x=5.02 \times 10^{-9}$

If the concentrations of both solutions are equal to or less than $5.02 \times 10^{-9} \mathrm{M}$, then there will be no precipitation of iron sulphide.