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lawyer [7]
2 years ago
13

When a mixture containing cations of Analytical Groups I–III is treated with H2S in acidic solution, which cations are expected

to precipitate? Analytical Groups I and II Analytical Group I only Analytical Group III only Analytical Groups II and III Analytical Group II only
Chemistry
2 answers:
Fantom [35]2 years ago
8 0

Answer:

Analytical Groups I and II

Explanation:

Precipitation reactions happen when anions and cations in aqueous solution mix together to form an insoluble ionic solid which is now referred to as a precipitate.

Whether or not a type of reaction like that takes place can be determined by utilizing the solubility principles for common ionic solids.

Tema [17]2 years ago
8 0

Answer:

Option D is the answer :

Analytical Groups II and III

Explanation:

Group II (Cu2+, Bi3+, Cd2+, Hg2+, As3+, Sb3+, Sn4+) cations produce very insoluble sulfides

(Ksp values less than 10-30) so they can be precipitated by low amounts of sulfide ion; this can be

achieved by adding an acidic solution of H2S.

Group III (Al3+, Cr3+, Fe3+, Zn2+, Ni2+, Co2+, Mn2+) cations produce slightly soluble sulfides

(Ksp values more than 10-20) so they can be precipitated by relatively high amounts of sulfide ion;

this can be achieved by adding a basic solution of H2S

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CaO + H2O -> Ca(OH)2
yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g

% yield :

\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%

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