How many grams of Ag2CO3 will precipitate when excess (NH4)2CO3 solution is added to 67.0 mL of 0.551 M AgNO3 solution?
2AgNO3(aq) + (NH4)2CO3(aq) 
Ag2CO3(s) + 2NH4NO3(aq)
First, we need to find how many moles of AgNO3 we are using. To do this, we use the volume and the molarity.
- We need to convert mL to liters. 67 mL is .067L
- We need to find how many moles are in .067 L of AgNO3 at 0.551 M. 0.551 M is the same as 0.551 mol/L.
- Solve for moles: 0.551 mol/L * 0.67 L = 0.36917 moles AgNO3.
Now that we know we have 0.36917 moles of AgNO3 , we can look at the rest of the problem. We are adding an excess of reactant to the reaction, so we will focus on the AgNO3 because it is our limiting reagent here. We need to find the mass of Ag2CO3, which you can do by hand or you can just google it. Using the latter method, we find that Ag2CO3 has a molar mass of 275.75 g/mol. Now, using molar ratios from our equation, we can solve!
2AgNO3(aq) + (NH4)2CO3(aq) Ag2CO3(s) + 2NH4NO3(aq)
0.36917mol AgNO3 *(1 mole Ag2CO3/2 mole AgNO3) * (275.75 g Ag2CO3/1 mol Ag2CO3) = 50.89 grams.
