how many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 64.0 mL of 0.554 M AgNO3 solution?
To find the number of grams of Ag2CO3 that will precipitate when excess Na2CO3 is added, we need to calculate the limiting reagent.
1. Start by writing the balanced equation for the reaction between AgNO3 and Na2CO3:
AgNO3 + Na2CO3 --> Ag2CO3 + 2NaNO3
2. Next, calculate the number of moles of AgNO3 present in the 64.0 mL of 0.554 M AgNO3 solution:
Moles of AgNO3 = Volume (L) x Concentration (M)
Moles of AgNO3 = 0.064 L x 0.554 M = 0.0354 mol
3. From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of Ag2CO3, so the number of moles of Ag2CO3 formed is also 0.0354 mol.
4. Now, we need to convert the moles of Ag2CO3 to grams. The molar mass of Ag2CO3 can be calculated as follows:
Ag: 2 x atomic mass of Ag = 2 x 107.87 g/mol = 215.74 g/mol
C: atomic mass of C = 12.01 g/mol
O: 3 x atomic mass of O = 3 x 16.00 g/mol = 48.00 g/mol
Total molar mass of Ag2CO3 = 215.74 + 12.01 + 48.00 = 275.75 g/mol
5. Convert the moles of Ag2CO3 to grams:
Grams of Ag2CO3 = Moles of Ag2CO3 x Molar mass of Ag2CO3
Grams of Ag2CO3 = 0.0354 mol x 275.75 g/mol = 9.77 g
Therefore, approximately 9.77 grams of Ag2CO3 will precipitate when excess Na2CO3 is added to 64.0 mL of 0.554 M AgNO3 solution.
To determine the number of grams of Ag2CO3 that will precipitate when Na2CO3 solution is added, we need to use stoichiometry and the concept of limiting reactants.
First, let's write the balanced equation for the reaction between AgNO3 and Na2CO3:
2 AgNO3 + Na2CO3 → Ag2CO3 + 2 NaNO3
From the balanced equation, we can see that the molar ratio between Ag2CO3 and AgNO3 is 1:2. This means that for every 2 moles of AgNO3, 1 mole of Ag2CO3 is produced.
Now let's calculate the moles of AgNO3 in the given volume:
Moles of AgNO3 = concentration (M) × volume (L)
Moles of AgNO3 = 0.554 M × 0.0640 L
Moles of AgNO3 = 0.035456 mol
Since the molar ratio between Ag2CO3 and AgNO3 is 1:2, the moles of Ag2CO3 formed will be half of the moles of AgNO3. Let's calculate the moles of Ag2CO3:
Moles of Ag2CO3 = 0.035456 mol / 2
Moles of Ag2CO3 = 0.017728 mol
Next, we need to calculate the molar mass of Ag2CO3. The molar mass of silver (Ag) is 107.87 g/mol, and the molar mass of carbonate (CO3) is 60.01 g/mol. Adding these together, we get:
Molar mass of Ag2CO3 = (2 × molar mass of Ag) + molar mass of CO3
Molar mass of Ag2CO3 = (2 × 107.87 g/mol) + 60.01 g/mol
Molar mass of Ag2CO3 = 275.75 g/mol
Finally, let's calculate the mass of Ag2CO3 that will precipitate:
Mass of Ag2CO3 = Moles of Ag2CO3 × Molar mass of Ag2CO3
Mass of Ag2CO3 = 0.017728 mol × 275.75 g/mol
Mass of Ag2CO3 = 4.8936 g
Therefore, approximately 4.8936 grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 64.0 mL of 0.554 M AgNO3 solution.
2AgNO3 + Na2CO3 ==> Ag2CO3 + 2NaNO3
mols AgNO3 = M x L = ?
Convert mols AgNO3 to mols Ag2CO3.
g Ag2CO3 = mols Ag2CO3 x molar mass.