A student wishes to prepare 1.000 g of silver chromate.
How many mL of .05000M silver nitrate solution should be added to 50.00 mL of .0700M potassium chromate solution assuming 100% yield
2AgNO3 + K2CrO4 ==> Ag2CrO4 + 2KNO3
mols K2CrO4 = M x L = ?
Convert to mols AgNO3. mols AgNO3 = 2*mols K2CrO4.
Then M AgNo3 = mols/L. You know mols and you M from the problem. Solve for L and convert to mL.
To determine the volume of silver nitrate solution needed to prepare 1.000 g of silver chromate, we need to calculate the number of moles of silver chromate required and then use stoichiometry to find the corresponding number of moles of silver nitrate.
1. Calculate the molar mass of silver chromate (Ag2CrO4):
- Ag (silver) has a molar mass of 107.87 g/mol
- Cr (chromium) has a molar mass of 52.00 g/mol
- O (oxygen) has a molar mass of 16.00 g/mol
So the molar mass of Ag2CrO4 is: (2 * 107.87) + 52.00 + (4 * 16.00) = 331.87 g/mol
2. Convert the given mass of silver chromate to moles:
Divide the given mass (1.000 g) by the molar mass (331.87 g/mol):
1.000 g / 331.87 g/mol = 0.003016 mol
3. Use the balanced chemical equation to determine the stoichiometric relationship between silver chromate and silver nitrate:
2 AgNO3 + K2CrO4 → Ag2CrO4 + 2 KNO3
From the balanced equation, we see that 2 moles of silver nitrate react to form 1 mole of silver chromate.
4. Calculate the number of moles of silver nitrate needed:
Since the stoichiometric ratio is 2:1 (silver nitrate to silver chromate), we need half the number of moles of silver chromate:
0.003016 mol / 2 = 0.001508 mol
5. Calculate the volume of the silver nitrate solution:
To find the volume, we can use the formula:
Volume (in liters) = Moles / Concentration
First, convert the given concentration of the silver nitrate solution from Molarity (M) to moles per liter (mol/L):
0.05000 M = 0.05000 mol/L
Then, rearrange the formula to solve for volume (in liters):
Volume (in liters) = Moles / Concentration
Volume (in liters) = 0.001508 mol / 0.05000 mol/L = 0.03016 L
Finally, convert the volume from liters to milliliters (mL):
0.03016 L * 1000 mL/L = 30.16 mL
Therefore, 30.16 mL of the 0.05000 M silver nitrate solution should be added to 50.00 mL of the 0.0700 M potassium chromate solution to prepare 1.000 g of silver chromate, assuming 100% yield.
To calculate the amount of silver nitrate solution needed, we can use the following balanced chemical equation:
AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
From the equation, we can see that 1 mole of silver nitrate (AgNO3) reacts with 1 mole of potassium chromate (K2CrO4) to produce 1 mole of silver chromate (Ag2CrO4).
First, we need to calculate the moles of potassium chromate (K2CrO4) in the 50.00 mL of .0700M solution:
Moles of K2CrO4 = concentration (M) x volume (L)
Moles of K2CrO4 = 0.0700 M x 0.05000 L
Moles of K2CrO4 = 0.0035 moles
According to the balanced equation, the ratio of moles of silver chromate (Ag2CrO4) to moles of potassium chromate (K2CrO4) is 1:1.
Therefore, we need 0.0035 moles of silver nitrate (AgNO3) to react with the potassium chromate (K2CrO4) in the solution.
Next, we need to calculate the volume (in liters) of the 0.05000M silver nitrate (AgNO3) solution required:
Moles of AgNO3 = 0.0035 moles
Molarity of AgNO3 = 0.05000 M
Volume of AgNO3 = Moles of AgNO3 / Molarity of AgNO3
Volume of AgNO3 = 0.0035 moles / 0.05000 M
Volume of AgNO3 = 0.07 L or 70 mL
Therefore, 70 mL of the 0.05000M silver nitrate solution should be added to 50.00 mL of the 0.0700M potassium chromate solution to prepare 1.000 g of silver chromate, assuming 100% yield.