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A. What is the molarity of sodium ions in a solution prepared by mixing 202.1 ml of 0.63 M sodium phosphate with 239 ml of 1.31 M sodium sulfide. Enter to 2 decimal places.



A. What is the molarity of sodium ions in a solution prepared by mixing 202.1 ml of 0.63 M sodium phosphate with 239 ml of 1.31 M sodium sulfide. Enter to 2 decimal places.

B. A 17.3 ml sample of a 1.49 M potassium chloride solution is mixed with 58.6 ml of a 1.34 M lead(II) nitrate solution and a precipitate forms. The solid is collected, dried, and found to have a mass of 0.5 g. What is the percent yield? Enter to 2 decimal places.

Solutions

Expert Solution

A) millimoles of Na+ from sodium phosphate = 3 x 202.1 x 0.63 = 382.0

     millimoles of Na+ from sodium sulfide = 2 x 239 x 1.31 = 626.2

molarity of sodium ion   = 626.2 + 382.0 / (202.1 + 239)

                                = 2.29 M

molarity of sodium ion = 2.29 M

B)

moles of KCl = 17.3 x 1.49 / 1000

                    = 0.0258

moles lead(II) nitrate = 58.6 x 1.34 / 1000

                               = 0.0785

2 KCl + Pb (NO3)2 --------------------------> PbCl2 + 2 KNO3

2                1

0.0258        0.0785

limiting reagent is KCl

so moles precipitate formed = 0.0258 / 2

                                         = 0.01425

mass of precipitate = 0.01425 x 278.1

                            = 3.96 g

percent yield = 0.5 x 100 / 3.96

percent yield = 12.62 %

                   

queen_honey_blossom answered 2 years ago
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