Clearly demonstrate which reactant salt would be limiting if the experiment was performed with solutions of equal concentrations. Assume 20 mL each of 0.5 M solutions, for instance. Show calculations ## Filtration Procedure### Figure 6.1Illustration of opened filter paper ready for placement into a filter funnel.### Method7. **Preparation** - Stir the reaction mixture thoroughly before beginning the filtration process. Avoid pouring the liquid directly onto the filter paper initially, as this may cause most of the solid particles to pass through the filter pores.8. **Filtration** - Pour the first sample through the filter paper. Use a rubber policeman or Teflon™ spatula to transfer the precipitate. Add no more than three 5 mL portions of warm wash water to rinse. - Carefully collect the precipitate on the filter paper. Post-filtration, remove and place the filter paper on a clean watch glass for drying until the next laboratory session.9. **Testing** - Using a plastic pipet, fill two centrifuge tubes halfway with the filtrate. Centrifuge if necessary, and label them tubes I and II. - To tube I, add two drops of BaCl₂ test reagent; to tube II, add two drops of Na₃PO₄ test reagent. An observable precipitation reaction should occur in one tube.10. **Secondary Filtration** - Discard the first filtrate into the waste container. Rinse the receiving flask thrice with deionized water, and repeat the filtration for the second sample as with the first.11. **Drying** - Remove the filter paper, unfold, and place on a watch glass. Dry the sample in the oven for approximately 15 minutes.12. **Measurement** - Record the mass of the dried filter paper and the collected precipitate.### Waste Disposal- **Liquid Waste**: Dispose of in the aqueous waste container.- **Solid Waste**: Dispose of in the solid waste container.### Clean-up- Rinse all glassware thoroughly with deionized water, dry, and return them to their proper storage. **Materials:**- Bunsen Burner- Ring stand- Wire mesh- 100 mL graduated cylinder (2)- 250 mL beakers (2)- 150 mL beaker- Test tubes (2)- Test tube rack- Glass stir rods (2)- Filtering funnel- Plastic transfer pipet- 250 mL Erlenmeyer flask**Note:** You will run two samples simultaneously. Perform the procedures for both samples. Obtain approximately the same volume of the same unknown samples for both trials.1. **Label** the graduated cylinders with labelling tape. One will be used for the BaCl₂ solution and the other for the Na₃PO₄ solution.2. **Obtain** approximately 20 mL each of an unknown BaCl₂ and Na₃PO₄ solutions. Record the volumes (to the correct number of significant figures) in your notebook.3. In the 250 mL beaker, **combine** the 20 mL samples of BaCl₂ and Na₃PO₄ solutions and 100 mL of deionized water. Stir each trial solution for 3 minutes. When finished stirring, remove the stir rod from the beaker, cover with a watch glass, and allow the precipitate to settle.4. While the precipitate is settling, set up the ring stand with wire mesh, light the Bunsen burner, and adjust the flame to a blue flame (with no traces of yellow). Place one sample on the wire mesh and digest the precipitate for 15 minutes. *Do not let the mixture come to a boil!* The goal is to simply gently heat the reaction mixture. Remove the first sample from the ring stand after 15 minutes and digest the second sample for 15 minutes.5. After both reactions have finished digesting, **heat** about 100 mL of deionized water to 70-80°C in the 150 mL beaker. This water will be used for washing the precipitate out of the beakers and for washing precipitate off of the stir rods.6. **Set up** the filtration apparatus. Obtain 2 sheets of fine porosity filter paper. Using pencil, write your initials and sample number on each paper. Fold each piece of filter paper into quarters. Then, open the filter as shown in Figure 6.1. Place the first filter

The given experiment is performed with 20 mL each of 0.5 M solutions of BaCl2 and Na3PO4.First of all, let us see the chemical reaction that occurs- 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl The above equation is balanced and we can see that 3 moles of BaCl2 reacts with 2 moles of Na3PO4. Now, the given moles of each solution = molarity x volume = 0.5 M x 0.020 L = 0.01 moles 2 moles of Na3PO4 ---> 3 moles of BaCl2 1 mole of Na3PO4 -----> 32 moles of BaCl2 0.01 mol of Na3PO4 -----> 32 x 0.01 = 0.015 mol of BaCl2 = 0.015 mol of BaCl2 required But we have only 0.01 mol of BaCl2 So, the limiting reagent is BaCl2.