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Alex787 [66]
2 years ago
11

Why can tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addit

ion of sodium hydroxide
Chemistry
1 answer:
devlian [24]2 years ago
7 0

Answer:

Tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide because they become soluble whereas copper (II) sulfide and bismuth (III) sulfide remain insoluble.

Explanation:

Sodium hydroxide is a  basic solution which is used as a precipitating agent for metallic ions in the laboratory.

<em>When a solution containing a mixture of the sulfides of the Group II cations,  antimony (III), copper (II), tin  (IV), and bismuth (III), is made basic by the addition of a base such as sodium hydroxide or ammonium hydroxide, the  sulfide ion concentration will increase. The sulfides of antimony (III) and tin (IV) will then become  soluble because antimony (III) and tin (IV) form stable complexes with sulfide, which are soluble in  water, while the sulfides of copper (II) and bismuth (III) do not.</em> The result is the dissolution of the antimony (III) sulfide  and tin (IV) sulfide, separating them from the copper (II) sulfide and the bismuth (III)  sulfide.

Sb₂S₃(s) + 3S²⁻(aq) ----> 2SbS₃³⁻(aq)

SnS₂(s) + S²⁻(aq) ----> SnS₃³⁻(aq)

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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
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Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

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According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

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Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

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Explanation:

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