The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $$H_2S$$ in the presence of conc. HCl (assuming 100% conversion) is

To find how many moles of $$As_2S_5$$ can be formed, we first change the given mass of arsenic acid $$H_3AsO_4$$ into moles. We use the formula $$n=\dfrac{m}{M}$$, where $$n$$ is the number of moles, $$m$$ is the mass in grams, and $$M$$ is the molar mass.

Molar mass of $$H_3AsO_4$$:

We have $$3$$ hydrogen atoms, $$1$$ arsenic atom, and $$4$$ oxygen atoms.

Hydrogen: $$3\times1\ \text{g mol}^{-1}=3\ \text{g mol}^{-1}$$

Arsenic: $$1\times75\ \text{g mol}^{-1}=75\ \text{g mol}^{-1}$$

Oxygen: $$4\times16\ \text{g mol}^{-1}=64\ \text{g mol}^{-1}$$

Adding them, $$M(H_3AsO_4)=3+75+64=142\ \text{g mol}^{-1}$$.

Now we calculate the moles of arsenic acid present:

$$n(H_3AsO_4)=\dfrac{35.5\ \text{g}}{142\ \text{g mol}^{-1}}=0.25\ \text{mol}$$

Next, we need the balanced chemical equation for the conversion of arsenic acid to arsenic pentasulphide in the presence of excess $$H_2S$$ (with conc. $$HCl$$ merely providing an acidic medium). The balanced equation is

$$2\,H_3AsO_4 + 5\,H_2S \rightarrow As_2S_5 + 8\,H_2O$$

From this equation we see that 2 moles of $$H_3AsO_4$$ give 1 mole of $$As_2S_5$$. Therefore, the mole ratio is $$2:1$$.

We have only $$0.25$$ mol of $$H_3AsO_4$$, so the moles of $$As_2S_5$$ formed will be

$$n(As_2S_5)=\dfrac{0.25\ \text{mol}}{2}=0.125\ \text{mol}$$

Thus, the amount of arsenic pentasulphide obtained is $$0.125\ \text{mol}$$.

Hence, the correct answer is Option D.