SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
{"Image DocumentName": "SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together", "Image ImageDescription": "Note: Integrative exercises require skills from earlier chapters as well as ones from the present chapter. A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) Calculate the theoretical yield, in grams, of the precipitate that forms. Solution (a) Potassium phosphate and silver nitrate are both ionic compounds. Potassium phosphate contains K+ and PO43\u2013 ions, so its chemical formula is K3PO4. Silver nitrate contains Ag+ and NO3\u2013 ions, so its chemical formula is AgNO3. Because both reactants are strong electrolytes, the solution contains K+, PO43\u2013, Ag+, and NO3\u2013 ions before the reaction occurs. According to the solubility guidelines in Table 4.1, Ag+ and PO43\u2013 form an insoluble compound, so Ag3PO4 will precipitate from the solution. In contrast, K+ and NO3\u2013 will remain in solution because KNO3 is water soluble. Thus, the balanced molecular equation for the reaction is. (b) To determine the limiting reactant, we must examine the number of moles of each reactant. \u2022 (Section 3.7) The number of moles of K3PO4 is calculated from the mass of the sample using the molar mass as a conversion factor. \u2022 (Section 3.4) The molar mass of K3PO4 is 3(39.1) (16.0) = g/mol. Converting milligrams to grams and then to moles, we have. We determine the number of moles of AgNO3. from the volume and molarity of the solution. \u2022 (Section 4.5) Converting milliliters to liters and then to moles, we have.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}