mols AgNO3 = L x M = 0.0700 x 0.692 = ?
Use the coefficients in the balanced equation to convert mols AgNO3 to mols Ag2CO3. That's
? mols AgNO3 x (1 mols Ag2CO3/2 mols AgNO3) = ? mols AgNO3 x 1/2 = ?
Then g Ag2CO3 = mols Ag2CO3 x molar mass Ag2CO3.
How many grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 70.0 mL of 0.692 M AgNO3 solution?
2AgNO3(aq) + K2CO3(aq) --> Ag2CO3(s) + 2KNO3(aq)
1 answer