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)
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.
To determine the amount of Ag2CO3 that will precipitate, you need to calculate the limiting reagent first. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To calculate the limiting reagent, we need to compare the stoichiometric ratios of the two reactants (AgNO3 and K2CO3) to the amounts given.
First, let's calculate the number of moles of AgNO3 in the solution:
moles AgNO3 = Molarity x Volume
moles AgNO3 = 0.692 mol/L x 0.0700 L
moles AgNO3 = 0.04844 mol
Since there is a 2:1 stoichiometric ratio between AgNO3 and Ag2CO3, we can calculate the number of moles of Ag2CO3 formed as follows:
moles Ag2CO3 = 0.04844 mol AgNO3 x (1 mol Ag2CO3 / 2 mol AgNO3)
moles Ag2CO3 = 0.02422 mol
Now, to find the mass of Ag2CO3 precipitated, we need to use the molar mass of Ag2CO3:
Molar mass of Ag2CO3 = (2 x atomic mass of Ag) + atomic mass of C + (3 x atomic mass of O)
Molar mass of Ag2CO3 = (2 x 107.87 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol)
Molar mass of Ag2CO3 = 275.77 g/mol
Finally, we can calculate the mass of Ag2CO3:
mass Ag2CO3 = moles Ag2CO3 x molar mass of Ag2CO3
mass Ag2CO3 = 0.02422 mol x 275.77 g/mol
mass Ag2CO3 = 6.69 g
Therefore, approximately 6.69 grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 70.0 mL of 0.692 M AgNO3 solution.
To determine the number of grams of Ag2CO3 that will precipitate, we need to first calculate the moles of AgNO3 in the solution and then use the stoichiometry of the balanced equation to convert moles of AgNO3 to moles of Ag2CO3. Finally, we can convert the moles of Ag2CO3 to grams.
Step 1: Calculate moles of AgNO3
Moles = Concentration (M) x Volume (L)
Moles of AgNO3 = 0.692 M x 0.0700 L
Moles of AgNO3 = 0.0484 moles
Step 2: Use stoichiometry to convert moles of AgNO3 to moles of Ag2CO3
According to the balanced equation:
2 moles of AgNO3 produce 1 mole of Ag2CO3
Therefore, 0.0484 moles of AgNO3 will produce (0.0484/2) moles of Ag2CO3
Step 3: Convert moles of Ag2CO3 to grams
Molar mass of Ag2CO3 = (2 x atomic mass of Ag) + atomic mass of C + (3 x atomic mass of O)
Molar mass of Ag2CO3 = (2 x 107.87 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol)
Molar mass of Ag2CO3 = 275.74 g/mol
Grams of Ag2CO3 = Moles x Molar mass
Grams of Ag2CO3 = (0.0484/2) moles x 275.74 g/mol
Calculate the result to find the final answer.