3. Copper (I) sulfide is insoluble in aqueous ammonia, while copper 0) sulfide d

Question

3. Copper (I) sulfide is insoluble in aqueous ammonia, while copper 0) sulfide dissolves to give a blue solution. Using the chemical fume hood,Group3 heated a small wire coil copper covered with sufficient powdered sulfur of inside a crucible. They continued heating until no more smoking is observed. A dull black brittle substance was left in the crucible. To a sample of the brittle substance inside a test tube, Group 3 added 2 ml of 6MNHdoil. The mixture remained unchanged after heating the test tube gently. Based on this observation, complete and balance the equation Why is it necessary to perform this experiment in the fume hood? Atest tube containing 3 ml of (PboNO) solution was added with 3 mL on and a white precipitate was formed. (a) Identify the type ofreaction, and (b) Write the complete and balanced equation and the chemical names of the reactants and products to support your answer in (a). 5. The equation below shows the precipitation reaction of copper nitrate and potassium iodide. To 2.0 ml of 0.50 McuoNosoh is added 50 ml of 0.5 M Ku solution. Calculate the theoretical mass of copper iodide (Cu) precipitate that is formed. 6. To deliver a 5.00 mL liquid sample most precisely, which piece of glassware would you use? (b) graduated cylinder (a) volumetric flask (c) glass dropper (c) volumetric pipet

Explanation / Answer

3. 8Cu + S8 -------> 8CuS CuS is insolublein ammonia.

This reaction has to be performed in fume hood beccaue on heating Suphur it produces SO2 which has a suffocating and irritating odour.

4. a. Double Displacement reaction.

Pb(NO3)2 + 2KCl -------> PbCl2 + 2KNO3

lead nitrate potassium chloride lead chloride (white ppt) potassium nitrate

5. 2 Cu(NO3)2   + 4KI -------> 2CuI + I2 + 4KNO3

2 mL 0.5 M Cu(NO3)2 = 2 x 0.5 = 1 mmol ( mmol = molarity x volume in mL)

5 mL 0.5 M KI = 5 x 0.5 = 2.5 mmol

1 equiv of Cu(NO3)2 will react with 2 equiv of KI

So the limiting reagent is Cu(NO3)2

1 equiv of Cu(NO3)2   will produce 1 equiv of CuI

Theoretical yield = 1 mmol CuI = 1 x 190.45 mg = 190.45 mg = 0.1904 g ( molar mass of CuI = 190.45 g/mol)

6. c. volumetric pipet using voluemtric pipet of capacity 5.0 ml , liquid sample can be taken precisely.

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