Question

# The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess ${H_2}S$ in the presence of conc. HCl (assuming 100% conversion) is:(A) 0.50mol(B) 0.125mol(C) 0.333mol(D) 0.25mol

Hint: First calculate the number of moles of arsenic acid involved here. Then using the reaction and the unitary method find out the number of moles of arsenic pentasulfide associated with it.

- We will begin with the reaction involved when arsenic acid is treated with excess ${H_2}S$ in the presence of conc. HCl and forms arsenic pentachloride is:
$2{H_3}As{O_4}\xrightarrow[{HCl}]{{{H_2}S}}A{s_2}{S_5}$
The question mentions that there is 100% conversion.
- Also the question mentions that the given weight of arsenic acid (${H_3}As{O_4}$) is = 35.5g
We can calculate its molecular weight on our own: Mol.wt = 3 + 75 + 64
= 142 g/mol
So, now we can find out the number of moles of ${H_3}As{O_4}$= given weight / mol.wt.
= 35.5 / 142
= 0.25 mol
- The reaction also tells us that from 2 moles of ${H_3}As{O_4}$ we can obtain 1 mole of $A{s_2}{S_5}$. So we will use the unitary method to find the moles of $A{s_2}{S_5}$ associated with 0.25 mol of ${H_3}As{O_4}$.
From 1 mol ${H_3}As{O_4}$ → $\frac{1}{2}$ mol of $A{s_2}{S_5}$ are obtained
So, from 0.25 mol ${H_3}As{O_4}$ → $\frac{1}{2} \times 0.25$ mol of $A{s_2}{S_5}$ are obtained
= 0.125 mol $A{s_2}{S_5}$
So, the correct option is: (B) 0.125mol.

Note: Completely balance the reaction between arsenic acid and arsenic pentachloride before solving. Arsenic acid is highly toxic and carcinogenic, thus it has limited applications. It can be used in making pesticides, wood preservatives, biocide, etc.