
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
Answer
583.8k+ views
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.
Complete step by step answer:
- 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.
Complete step by step answer:
- 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.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

