Question

a) How many grams of the insoluble product are formed after mixing 187 mL of 0.611 M solution of potassium phosphate with 90.9 mL of 0.492 M solution of calcium sulfate?

b) What is the percent yield if 1.39 g of chlorine were obtained by reacting 17.04 g of elemental iodine with 30.2 mL of 1.68 M sodium chloride?

c) How many grams of a soluble compound are dissolved in water, if 797 mL of a 1.50 M solution of the compound are prepared in a beaker. Assume the molar mass of the soluble compound to be 118.34 g/mol.

Answer 1

@ Mole) Kapag ع Tooo (Moley Cason .: )m Litee Ca (pon) 2 (3) 14.624 ngon Molarity x rolume (Litre) 187 Litre X 0.611 Molar. 0.114257 mol 0.49 M X (go.9 Litre 100 o o.0447 228 mol 4 Корея + 3 Casoy 3 Kg 504 + I 0.114257 mol 0.0447228 mo с -0.0447228X2 0.044 7228 to.0447228 0.044722844 릉 3 0.044 7228 0.014 30 76. 0.0380 0.044 7 2 28xL- o.0149 676 mole Man Molex Molar Mars = 0.0149076 mole x 310.1887 me 3 4 (Moles Cag fonda 28x16 (3 Mole Mulanity & Volume. Mole 1.50 mol Litre х 797 Litre 1000 1.1955 mole. (Mars) = Holex Molar Mars dissolved 1.1955 mole x 118.34 ilmol 141-475 am

(Mole) 30.9 L x NAC 1.68 mol Litre 6.050736 mol (Moles 11 17.04 am IQ = 0.067137 253.8089 g/mol H + 9 Nach & NaI t cla I 0.067137 0:050736 с -0.056736 -0.050736 mol QC0.0 50736)mol 0.0507 36 mol f 0.101474 0.041757 0.050736 (Mars) cia Mole) x Molar Mars = 0.050 736 mole X 70.906 mol (Flars) cia il 3.597486 gm [ Theoritical yield Mare % yield Actual yield The ritical yield X 100 1-39 g x 100% 3.5974868 38.6382