solve this problem: 5.05 g of silver nitrate are dissolved in 50 ml of deionized
ID: 678511 • Letter: S
Question
solve this problem: 5.05 g of silver nitrate are dissolved in 50 ml of deionizedwater. 5.50 grams of barium chloride are dissolved in 50 ml ofdeionized water in another beaker. The solutions are mixed together and a prcipitate forms. a) Write the balanced equasion for the reaction. b) Determine the identity of the precipitate. c) Determine the limiting factor and the excess reagent. d) Determine the mass of precipitate that would beformed. solve this problem: 5.05 g of silver nitrate are dissolved in 50 ml of deionizedwater. 5.50 grams of barium chloride are dissolved in 50 ml ofdeionized water in another beaker. The solutions are mixed together and a prcipitate forms. a) Write the balanced equasion for the reaction. b) Determine the identity of the precipitate. c) Determine the limiting factor and the excess reagent. d) Determine the mass of precipitate that would beformed.Explanation / Answer
We Know that : a ) AgNO3 + water ------> AgNO3 (aq.) BaCl2 + water -------> BaCl2 (aq.) 2 AgNO3(aq.) + BaCl2(aq.) --------> 2 AgCl + Ba (NO3)2 b ) ThePrecipitate obtained is due to AgCl i.e white ppt. c ) From the above equation it is clear that : AgNO3 acts as limiting reagent and BaCl2 is excess reagent. d ) The mass of the Precipitate formed is 0.032 moles x 179 gm /mole = 5.728 gm. b ) ThePrecipitate obtained is due to AgCl i.e white ppt. c ) From the above equation it is clear that : AgNO3 acts as limiting reagent and BaCl2 is excess reagent. d ) The mass of the Precipitate formed is 0.032 moles x 179 gm /mole = 5.728 gm.Related Questions
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