The treatment of an aqueous solution of 3.74 g of Cu(NO$$_3$$)$$_2$$ with excess KI results in a brown solution along with the formation of a precipitate. Passing H$$_2$$S through this brown solution gives another precipitate X. The amount of X (in g) is ______.

[Given : Atomic mass of H = 1, N = 14, O = 16, S = 32, K = 39, Cu = 63, I = 127]

Molar mass of $$Cu(NO_3)_2 = 63 + 2(14) + 6(16) = 187 \text{ g mol}^{-1}$$.

Moles of $$Cu(NO_3)_2$$ taken $$n_{Cu^{2+}} = \frac{3.74 \text{ g}}{187 \text{ g mol}^{-1}} = 0.020 \text{ mol}$$.

With excess $$KI$$ the reaction is $$2\,Cu^{2+} + 4\,I^- \;\longrightarrow\; 2\,CuI(s) + I_2(aq)$$.

From the stoichiometry, $$2$$ mol of $$Cu^{2+}$$ give $$1$$ mol of $$I_2$$. Hence moles of $$I_2$$ formed $$n_{I_2} = \frac{0.020}{2} = 0.010 \text{ mol}$$.

The brown colour of the filtrate is due to this $$I_2$$. Passing $$H_2S$$ through it: $$I_2 + H_2S \;\longrightarrow\; 2\,HI + S(s)$$.

Here, $$1$$ mol of $$I_2$$ produces $$1$$ mol of elemental sulfur $$S$$. Therefore, moles of precipitated sulfur $$n_S = 0.010 \text{ mol}$$.

Mass of sulfur obtained $$m_S = n_S \times M_S = 0.010 \times 32 = 0.32 \text{ g}$$.

Hence, the amount of precipitate X is 0.32 g.