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Some Basic Concepts of Chemistry question

2016 · 9 Apr · Shift 1 · Q21
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Some Basic Concepts of Chemistry question

2016 · 9 Apr · Shift 1 · Q21

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2SH_2SH2​S in the presence of conc. HCl ( assuming 100% conversion) is :
  1. A
    0.50 mol
  2. B
    0.25 mol
  3. C
    0.125 mol
  4. D
    0.333 mol
View written solutionFree

Correct answer: C

  1. Write the reaction

Arsenic acid is H3AsO4\mathrm{H_3AsO_4}H3​AsO4​. With excess H2S\mathrm{H_2S}H2​S in acidic medium, it forms arsenic pentasulphide:

2H3AsO4+5H2S→As2S5+8H2O2\mathrm{H_3AsO_4} + 5\mathrm{H_2S} \rightarrow \mathrm{As_2S_5} + 8\mathrm{H_2O}2H3​AsO4​+5H2​S→As2​S5​+8H2​O

So,

2 mol H3AsO4→1 mol As2S52\text{ mol } \mathrm{H_3AsO_4} \rightarrow 1\text{ mol } \mathrm{As_2S_5}2 mol H3​AsO4​→1 mol As2​S5​

  1. Calculate moles of arsenic acid

Molar mass of H3AsO4\mathrm{H_3AsO_4}H3​AsO4​:

3(1)+75+4(16)=3+75+64=142 g mol−13(1) + 75 + 4(16) = 3 + 75 + 64 = 142\,\mathrm{g\,mol^{-1}}3(1)+75+4(16)=3+75+64=142gmol−1

Given mass = 35.5 g35.5\,\mathrm{g}35.5g

n(H3AsO4)=35.5142=0.25 moln(\mathrm{H_3AsO_4}) = \frac{35.5}{142} = 0.25\,\mathrm{mol}n(H3​AsO4​)=14235.5​=0.25mol

  1. Use stoichiometric ratio

From the reaction:

2 mol H3AsO4→1 mol As2S52\text{ mol } \mathrm{H_3AsO_4} \rightarrow 1\text{ mol } \mathrm{As_2S_5}2 mol H3​AsO4​→1 mol As2​S5​

Therefore,

0.25 mol H3AsO4→0.252=0.125 mol As2S50.25\text{ mol } \mathrm{H_3AsO_4} \rightarrow \frac{0.25}{2} = 0.125\text{ mol } \mathrm{As_2S_5}0.25 mol H3​AsO4​→20.25​=0.125 mol As2​S5​

  1. Check options
  • A: 0.500.500.50 mol ×\times× incorrect
  • B: 0.250.250.25 mol ×\times× incorrect
  • C: 0.1250.1250.125 mol ✓\checkmark✓
  • D: 0.3330.3330.333 mol ×\times× incorrect

Hence, the correct option is C.

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