calculate the mass of solid NaCl that must be added to 1.50L of a 0.100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl. calculate the mass of AgCl formed.
AgNO3 + NaCl ==> NaNO3 + AgCl
mols AgNO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols NaCl. Then grams NaCl needed = mols NaCl x molar mass NaCl.
Use the same process for grams AgCl formed.
Please help me
To calculate the mass of solid NaCl that must be added, we need to determine the moles of Ag+ ions in the 1.50L of the AgNO3 solution, and then use the stoichiometry between Ag+ and NaCl to find the corresponding moles of NaCl.
Step 1: Calculate moles of Ag+ ions in the solution.
Molarity (M) = moles/volume (L)
0.100 M = moles/1.50 L
Moles of Ag+ ions = 0.100 M × 1.50 L = 0.150 moles
Step 2: Use stoichiometry to find moles of NaCl.
From the balanced equation:
AgNO3 + NaCl -> AgCl + NaNO3
Based on the equation, the stoichiometry between Ag+ and NaCl is 1:1, so the moles of Ag+ ions will be equal to the moles of NaCl.
So, moles of NaCl = 0.150 moles
Step 3: Calculate the mass of NaCl.
Molar mass of NaCl = atomic mass of Na + atomic mass of Cl
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Mass of NaCl = moles × molar mass = 0.150 moles × 58.44 g/mol = 8.77 grams
Therefore, 8.77 grams of solid NaCl must be added to precipitate all the Ag+ ions in the form of AgCl.
To calculate the mass of AgCl formed, we can use the same stoichiometry:
From the balanced equation:
AgNO3 + NaCl -> AgCl + NaNO3
Since the stoichiometry between AgNO3 and AgCl is 1:1, the moles of AgCl formed will also be 0.150 moles.
Mass of AgCl = moles × molar mass of AgCl
Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl = 0.150 moles × 143.32 g/mol = 21.50 grams
Therefore, the mass of AgCl formed will be 21.50 grams.
To determine the mass of solid NaCl that must be added, we need to calculate the number of moles of Ag+ ions in the 1.50L of 0.100M AgNO3 solution and then use stoichiometry to find the number of moles of NaCl required.
Step 1: Calculate the moles of Ag+ ions in the AgNO3 solution.
Molarity (M) is defined as moles of solute per liter of solution. We have a 0.100M AgNO3 solution with a volume of 1.50L. Using the formula:
moles = Molarity × volume
moles of AgNO3 = 0.100M × 1.50L
moles of AgNO3 = 0.150 mol
Since AgNO3 dissociates to produce one Ag+ ion, the moles of Ag+ ions is also 0.150 mol.
Step 2: Use stoichiometry to calculate the moles of NaCl required.
The balanced equation for the reaction of Ag+ ions with Cl- ions to form AgCl is:
Ag+ + Cl- -> AgCl
From the balanced equation, we can see that one mole of Ag+ ions (or AgNO3) reacts with one mole of Cl- ions (from NaCl) to form one mole of AgCl.
moles of NaCl = moles of Ag+ ions
moles of NaCl = 0.150 mol
Step 3: Calculate the mass of NaCl required.
To find the mass of NaCl, we need to use the molar mass of NaCl (sodium chloride). The molar mass of NaCl is 58.44 g/mol.
mass = moles × molar mass
mass of NaCl = 0.150 mol × 58.44 g/mol
mass of NaCl = 8.766 g (rounded to three decimal places)
So, the mass of solid NaCl that must be added to the solution is 8.766 grams.
Step 4: Calculate the mass of AgCl formed.
From the stoichiometry of the balanced equation, we can see that one mole of AgCl is formed per mole of Ag+ ions.
mass of AgCl = moles of Ag+ ions × molar mass of AgCl
mass of AgCl = 0.150 mol × (107.87 g/mol)
mass of AgCl = 16.18 g (rounded to two decimal places)
Therefore, the mass of AgCl formed is 16.18 grams.