CaCl2+AgNO3>>2AgCl + ....
Ok, you first figure the moles of silver chloride form 2.9grams.
Then you had half that number of moles of calcium chloride in solutions, so
MolarityCaCl2=molesCaCl2/.025
If an excess of silver nitrate is combined with 25.0 mL of calcium chloride and produces 2.90 g of precipitate, what is the molarity of calcium chloride that was used?
1 answer
1 answer