A 15.5 mL sample of 0.545 M lead(II) nitrate soution is mixed with 16.5 mL of...

A 15.5 mL sample of 0.545 M lead(II) nitrate soution is mixed with 16.5 mL of 0.500 M potassium iodide solution and the precipitation reaction in [ 2KI+pb(NO3)2--->PbI2+2KNO3) ]. Determine the precipitate using the solubility rules. The solid is collected, dried, and found to have a mass of 1.79 g. Determine the limiting reactant, the theoretical yield and the percent yield.

Expert Solution

queen_honey_blossom answered 2 years ago

Part 1. A 13 ml sample of a 1.38 M potassium chloride solution is mixed with 52.9 ml of a 1.43 M lead(II) nitrate solution and a precipitate forms.

Part 1. A 13 ml sample of a 1.38 M potassium chloride solution is mixed with 52.9 ml of a 1.43 M lead(II) nitrate solution and a precipitate forms. The solid is collected, dried, and found to have a mass of 0.9 g. What is the percent yield? Enter to 2 decimal places.Part 2. What is the net ionic equation of barium hydroxide + hydrofluoric acid?

A 450.0 mL sample of a 0.295 M solution of silver nitrate is mixed with 400.0...

A 450.0 mL sample of a 0.295 M solution of silver nitrate is mixed with 400.0 mL of 0.200 M calcium chloride. What is the concentration of Cl- in solution after the reaction is complete?

when a solution of lead(ii) nitrate is mixed with a solution of potassium chromate, a yellow...

when a solution of lead(ii) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation: Pb(NO3)2 (aq) + K2CrO4 (aq) -> 2 KNO3 (aq) + PbCrO4. Volume of 0.105 M lead(ii) nitrate react with 100.0 ml of 0.120 M potassium chromate. What mass of PbCrO4 will be formed?

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M...

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M sodium sulfate (Na2SO4, 142 g/mol) are mixed together. (I) Write the overall balanced equation. Include the states of each species. (II) Write the overall (or total) ionic equation for the reaction. Identify any spectator ions. Include states of each species. (III) Write the net ionic equation. Include states of each species. (IV) Determine the limiting reactant and the theoretical yield of the non-aqueous product....

Consider mixing an excess of lead(II) nitrate (aq) with 200.0 mL of 0.400 M sodium chloride....

Consider mixing an excess of lead(II) nitrate (aq) with 200.0 mL of 0.400 M sodium chloride. Determine the mass of solid formed, assuming a complete reaction What volume (in mL) of 0.800 M lead(II) nitrate must you add to make sure you make the mass of product you calculated in problem 3? A 0.1044 g sample of the salt MCl was dissolved in water (making it in the aqueous phase). An excess of AgNO3 (aq) was added, precipitating 0.0889 g...

A 160.0 mL solution of 2.813 M strontium nitrate is mixed with 215.0 mL of a...

A 160.0 mL solution of 2.813 M strontium nitrate is mixed with 215.0 mL of a 2.900 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. Mass = 39.2 g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration. [Na+] = ? [Sr2+]=0.37 M [F-]= 0 M [NO3-]= ?

In a laboratory experiment, 2.0 g of aqueous lead (II) nitrate was mixed with 2.0 g...

In a laboratory experiment, 2.0 g of aqueous lead (II) nitrate was mixed with 2.0 g of aqueous sodium chloride. Upon completion of the experiment, 0.65 g of lead (II) chloride were obtained. a) Provide the balanced equation. b) What is the limiting reactant? c) What is the theoretical yield? d) What is the percent yield?

400. mL of 0.5875 M aqueous silver nitrate are mixed with 500. mL of 0.290 M...

400. mL of 0.5875 M aqueous silver nitrate are mixed with 500. mL of 0.290 M sodium chromate. a) What is the mass of the precipitate? b) What is the concentration of each of the ions remaining in solution? Indicate any assumptions or approximations.

11. A 150.0 mL solution of 2.713 M strontium nitrate is mixed with 220.0 mL of...

11. A 150.0 mL solution of 2.713 M strontium nitrate is mixed with 220.0 mL of a 2.401 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [Na+] M [NO−3] M [Sr2+] M [F−] M

Aqueous solutions of sodium sulfate and lead(II) nitrate are mixed and a white solid forms. What...

Aqueous solutions of sodium sulfate and lead(II) nitrate are mixed and a white solid forms. What is the formula of the solid? Click in the answer box to open the symbol palette.

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