Please show me how to do this step by step I would appreciate it
We add excess Na2CrO4 solution to 42.0 mL
of a solution of silver nitrate (AgNO3) to form
insoluble solid Ag2CrO4. When it has been
dried and weighed, the mass of Ag2CrO4 is
found to be 0.520 grams. What is the molarity
of the AgNO3 solution?
Answer in units of M.
M AgNO3 = grams AgNO3/L solution = ?
M AgNO3 = grams AgNO3/0.042 = ?
Therefore, to find M AgNO3 we must find grams AgNO3. We have the volume in L already. The Ag2CrO4 gives us a way to find the mass AgNO3 we started with.
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NanO3
mols Ag2CrO4 = grams/molar mass = ?
mols Ag2CrO4 = 0.520/about 332 = about 0.00157.
Using the coefficients in the balanced equation, convert mols Ag2CrO4 to mols AgNO3. That is
0.00157 x (2 mol AgNO3/1 mol Ag2CrO4) = 0.00157 x 2/1 = about 0.00313 mols AgNO3.
Convert mols AgNO3 to grams. g = mols AgNO3 x molar mass AgNO3. Then substitute that number (in grams) into the M equation at the top and solve for M. Note that I've estimated here and there so you need to go through and recalculate each step more accurately than I did.
To find the molarity of the AgNO3 solution, we can use the following steps:
Step 1: Write the balanced chemical equation for the reaction.
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
Step 2: Determine the number of moles of Ag2CrO4.
The molar mass of Ag2CrO4 is:
(2 x atomic mass of Ag) + atomic mass of Cr + (4 x atomic mass of O)
= (2 x 107.87 g/mol) + 52.00 g/mol + (4 x 16.00 g/mol)
= 331.87 g/mol
Given the mass of Ag2CrO4 is 0.520 grams, we can calculate the number of moles:
moles of Ag2CrO4 = mass of Ag2CrO4 / molar mass of Ag2CrO4
= 0.520 g / 331.87 g/mol
= 0.001567 mol
Step 3: Determine the number of moles of AgNO3.
From the balanced chemical equation, we can see that the stoichiometric ratio between Ag2CrO4 and AgNO3 is 2:2. This means that the number of moles of AgNO3 is the same as the number of moles of Ag2CrO4.
Therefore, the number of moles of AgNO3 is also 0.001567 mol.
Step 4: Calculate the volume of the AgNO3 solution.
We are given a volume of 42.0 mL for the AgNO3 solution.
Step 5: Convert the volume to liters.
volume of AgNO3 solution = 42.0 mL = 0.0420 L
Step 6: Calculate the molarity of the AgNO3 solution.
Molarity (M) = moles of solute / volume of solution in liters
= 0.001567 mol / 0.0420 L
= 0.03726 M
Therefore, the molarity of the AgNO3 solution is 0.03726 M.
To find the molarity of the AgNO3 solution, we need to use stoichiometry and the concept of molarity. Here's the step-by-step method:
Step 1: Write the balanced chemical equation for the reaction:
2 AgNO3 + Na2CrO4 -> Ag2CrO4 + 2 NaNO3
From the equation, we can see that the ratio between AgNO3 and Ag2CrO4 is 2:1.
Step 2: Calculate the number of moles of Ag2CrO4:
Given the mass of Ag2CrO4 is 0.520 grams, we need to convert it to moles using the molar mass of Ag2CrO4. The molar mass of Ag2CrO4 is:
(2 x Atomic mass of Ag) + Atomic mass of Cr + (4 x Atomic mass of O)
The atomic mass of Ag = 107.87 g/mol
The atomic mass of Cr = 51.996 g/mol
The atomic mass of O = 16.00 g/mol
Calculating the molar mass: (2 x 107.87) + 51.996 + (4 x 16.00) = 331.790 g/mol
Now, we can calculate the number of moles of Ag2CrO4:
Moles = Mass / Molar mass
Moles = 0.520 g / 331.790 g/mol
Step 3: Determine the number of moles of AgNO3:
Since the ratio between Ag2CrO4 and AgNO3 is 2:1, the number of moles of AgNO3 is half of the number of moles of Ag2CrO4. Therefore, we can write:
Moles (AgNO3) = Moles (Ag2CrO4) / 2
Step 4: Calculate the volume of the AgNO3 solution:
We are given the volume of the Na2CrO4 solution that was added to the AgNO3 solution, which is 42.0 mL.
Since the volume of AgNO3 solution is the same as the volume of Na2CrO4 solution, the volume of AgNO3 solution is 42.0 mL.
Step 5: Calculate the molarity of AgNO3:
Molarity (AgNO3) = Moles (AgNO3) / Volume (AgNO3)
Molarity (AgNO3) = Moles (AgNO3) / 42.0 mL
Remember that the volume should be converted to liters because the unit of molarity is mol/L.
To convert mL to L, divide the volume by 1000:
Volume (AgNO3) = 42.0 mL / 1000 mL/L
Now, substitute the values into the equation:
Molarity (AgNO3) = Moles (AgNO3) / Volume (AgNO3)
Molarity (AgNO3) = (Moles (Ag2CrO4) / 2) / (42.0 mL / 1000 mL/L)
Finally, calculate the value:
Molarity (AgNO3) = (0.520 g / 331.790 g/mol) / (42.0 mL / 1000 mL/L)
Molarity (AgNO3) = (0.001568 mol) / (0.042 L)
Molarity (AgNO3) ≈ 0.037 mol/L or 0.037 M
Therefore, the molarity of the AgNO3 solution is approximately 0.037 M.