1. What happened when you added test solution to the test tube filled with only

Question

1.What happened when you added test solution to the test tube filled with only copper sulfate solution?

2. What happened when you added test solution to the test tube filled with only sodium sulfide solution?

3. Which of the following contains an excess of CuSO4, or NaS, or neither? (excluding Test tubes 1 and 7)
Test tube 2: 5 mL CuSO4 and 1 mL NaS
Test tube 3: 4 mL CuSO4 and 2 mL NaS
Test tube 4: 3 mL CuSO4 and 3 mL NaS
Test tube 5: 2 mL CuSO4 and 4 mL NaS
Test tube 6: 1 mL CuSO4 and 5 mL NaS
test solutions used: 6M NH3(20mL) and 0.4 ZnSO4

4. Which test tube had the largest amount of precipitate? Record the ratio of CuSO4 to Na2S used. What is indicated by the maximum amount of precipitate?

5. Did the test solution react with the supernatant from the test tube that had the largest amount of precipitate? What can you infer about the molar ratio of this reaction?

6. Explain how excess reagents and the amount of precipitate is formed?

Explanation / Answer

1. When test solution is added to CuSO4 solution,

dark blue coloration is developed due to the formation of Cu(OH)2 species

2. When test solution was added to sodium sulfide solution,

(NH4)2S(aq) and ZnS(s) are formed

3. Excess reagent

test tube 2 : CuSO4

test tube 3 : CuSO4

test tube 4 : sodium sulfide

test tube 5 : sodium sulfide

test tube 6 : sodium sulfide

4. The test tube with largest amount of precipitate = test tube 4

This has largest amounts of CuSO4 and Na2S ratio. formation of white precipitate indicates this state of precipitate. Reagents are presen in 1 :1 ratio.

5. the test solution did not react with supernatant of the solution with largest amount of precipitate. This tells us that mole ratio of CuSO4 to Na2S is 1 : 1, so all has reacted to form precipitate and thus no reagent is available to react with test solution.

6. Excess reagent as in case of test tubes 2 and 3 are CuSO4 and in test tubes 5 and 6 as sodium sulfide. The amount of precipitate formed is calculated from the moles of limiting reactant. The amount of precipitate formed is equal to moles of limiitng reactant, leaving behind exces reacgent in solution.

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