A 55.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs:
The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, the theoretical yield, and the percent yield.
Solution: Here, we are going to identify which of the given pairs of solution has higher concentration of iodide ion. Step1: a) 1 molecule of BaI dissociates in aqueous solution to give 1 Ba ion and 2 I ions. 2+ - 2 - Therefore, 0.10 M BaI will g2e (2 x 0.10 = 0.20) M I ions in solution. Similarly, 1 molecule of KI dissociates in aqueous solution to give one K ion and one I ion. + - - Therefore, 0.25 M KI will produce 0.25 M I ions in solution. - Thus, 0.25 M KI solution has higher concentration of I ions. Step2: b) Number of moles of any compound in solution is given by: Number of moles of solute = Volume of solution in L X Molarity solute Therefore, number of moles of KI = (100 / 1000 L) X 0.10 M = 0.1 L X 0.10 M = 0.01 moles Similarly, number of moles of ZnI = (200 / 1000 L) X 0.040 M 2 = 0.2 L X 0.040 M =0.008 moles. Now, 1 mole of KIwill contain 1 mole of potassium ion. Therefore, 0.01 moles of KIwill contin 0.01 moles of potassium ions. Similarly, 0.008 moles of ZnI will contain (2 x 0.008 = 0.016) moles of potassium ions. 2 - Thus, 200 ml of 0.040 M Znl solution 2ll have higher concentration of I ions. Step3: c) Molarity of NaI solution is given by, Molarity = number of moles of NaI / Volume of solution in litres Mass of NaI dissolved = 145 g Molar mass of NaI = 149.8942 g Therefore, number of moles of NaI = 145 / 149.8942 = 0.967 moles Volume of solution in litres = 150/ 1000 L = 0.15 L Therefore, molarity = 0.967 / 0.15 = 6.45 M Now, 3.2 M HI will produce 3.2 M I ions and similarly - 6.45 M NaI will produce 6.45 M I ions. - - Thus, the NaI solution will have higher concentration of I ions. -----------------