The Fools Gold

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.20 g of sodium carbonate is mixed with one containing 4.57 of silver nitrate.

1- How many grams of silver nitrate are present after the reaction is complete?

2- How many grams of sodium nitrate are present after the reaction is complete?

Na2CO3 + 2 AgNO3 = Ag2CO3 + 2 NaNO3

moles Na2CO3 = 3.20 g / 105.988 g/mol= 0.0302
moles AgNO3 = 4.57 g / 169.87 g/mol=0.0269 => Limiting reactant

moles Na2CO3 required = 0.0269/2 =0.0135
moles Na2CO3 in excess = 0.0302 - 0.0135 = 0.0167
mass Na2CO3 in excess = 0.0167 x 105.988 = 1.77 g

AgNO3 is not present after the reaction is complete

One Response

1. Dr.A Says:
   February 17th, 2010 at 8:44 pm

   Na2CO3 + 2 AgNO3 = Ag2CO3 + 2 NaNO3

   moles Na2CO3 = 3.20 g / 105.988 g/mol= 0.0302
   moles AgNO3 = 4.57 g / 169.87 g/mol=0.0269 => Limiting reactant

   moles Na2CO3 required = 0.0269/2 =0.0135
   moles Na2CO3 in excess = 0.0302 - 0.0135 = 0.0167
   mass Na2CO3 in excess = 0.0167 x 105.988 = 1.77 g

   AgNO3 is not present after the reaction is complete
   References :

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