Question

Questions are after the procedures. Procedure 1) For Trial 1, use a graduated cylinder to measure...

Questions are after the procedures.

Procedure

1) For Trial 1, use a graduated cylinder to measure 5 mL of 1 M sodium sulfate (Na2SO4). Place the sodium sulfate in a beaker. Using a clean graduated cylinder, add 5 mL of 1 M calcium chloride (CaCl2) to the beaker. Stir to mix well.

2) Using two Büchner funnels as in the diagram below, set up a funnel system with a safety flask to protect the water system from contaminants.

3) Filter the solution and solid from Procedure 1 by suction through filter paper in the Büchner funnel. After drawing air through the calcium sulfate (CaSO4) precipitate for 5 min- utes, weigh the precipitate and filter paper on a watch glass. Record this mass on the Report Sheet.

4) Divide the filtered liquid from the filter flask into two beakers. Add 1 mL of 1 M calcium nitrate (Ca(NO3)2) to one beaker. Add 1 mL of 1 M copper(II) sulfate (CuSO4) to the second beaker. Notice the small amount of precipitation.

5) For Trial 2, use a graduated cylinder to measure 10 mL of 1 M sodium sulfate (Na2SO4). Place the sodium sulfate in a beaker. Using a clean graduated cylinder, add 10 mL of 1 M calcium chloride (CaCl2) to the beaker. Stir to mix well.

6) Filter the solution and solid from Procedure 5 by suction through filter paper in a Büchner funnel. After drawing air through the calcium sulfate (CaSO4) precipitate for 5 min- utes, weigh the precipitate and filter paper on a watch glass. Record this mass on the Report Sheet.

7) Divide the filtered liquid from the filter flask into two beakers. Add 1 mL of 1 M calcium nitrate (Ca(NO3)2) to one beaker. Add 1 mL of 1 M copper(II) sulfate (CuSO4) to the second beaker. Notice the small amount of precipitation.

8) For Trail 3, repeat Procedures 5, 6, and 7 but use 20 mL of 1 M sodium sulfate. In Procedure 7, take a mental note of the amount of precipitate generated from adding calcium nitrate.

9) For trial 4, repeat Procedures 5, 6, and 7 using 20 mL of 1 M calcium chloride with 10 mL of 1 M sodium sulfate. In Procedure 7, take a mental note of the amount of precipitate generated from the addition of copper sulfate.

10) Pour together all the calcium and sulfate solutions. Filter the calcium sulfate precipitate. Discard all precipitates in a container to be buried in a landfill unless directed to do otherwise by your instructor. Pour the remaining solutions down the drain with lots of water or dispose of as directed.

Questions are below. Some questions require the procedures, so I posted them above.

1. What remains in solution at the end of Procedure 1?

2. If sodium ions and chloride ions were left in solution, how could they be recovered?

3. If the reaction called for 1M chloride ion with 1 M calcium ion, would you still use 10 mL of each solution for a complete reaction? Explain your reasoning.

4. If it was determined that the water supply for a city had a minute level of lead ion contamination, the water treatment plant would likely attempt to eliminate the lead by precipitating it. Would they want the negative ion they add to the water or the existing lead ion to be the limiting reactant? Explain your reasoning.

5. How does knowing the correct number of moles to exactly react relate to the second principle of green chemistry—atom economy—as listed in Hill’s test?

Homework Answers

Answer #1

The reaction in procedure 1 is as follows;

Na2SO4 + CaCl2 = CaSO4 + 2NaCl

Sodium Chloride , NaCl remains dissolved in the solution (filtrate)

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2. NaCl could be recovered by evaporating the solution.

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3. The formula of Calsium chloride is CaCl2. This means that the salt contain double the amount of Chloride ion than Calcium ion. So, for a complete reaction between Ca and Chloride ion ratio of the volume should be 1:2.

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4. lead ion should be the limiting reagent. As, limiting reagent is completely revoved in a reaction.

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5. If correct number of moles are used, the reaction will be stoichiometric. It means that there will be maximum amount of product formation and also , reactant will not be in excess.

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