Approximately how many mL of 5% BaCl22H2O solution would be required to precipitate all the sulfate if we assume that your samples are pure sodium sulfate? Assume that the density of the barium chloride solution is 1.00 g/mL
How many g Na2SO4 do you have?
562.0 grams
To determine the number of mL of 5% BaCl22H2O solution required, we need to calculate the amount of BaCl22H2O needed to precipitate all the sulfate.
We can start by calculating the molecular weight of BaCl22H2O:
Ba = 137.33 g/mol
Cl2 = 2 * 35.45 g/mol = 70.90 g/mol
H2O = 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
Total molecular weight of BaCl22H2O = 137.33 g/mol + 70.90 g/mol + 18.02 g/mol = 226.25 g/mol
Next, we calculate the amount of BaCl22H2O needed to react with the sulfate. One mole of BaCl22H2O reacts with one mole of sulfate (Na2SO4):
1 mole BaCl22H2O : 1 mole Na2SO4
So, the amount of BaCl22H2O needed is equal to the amount of Na2SO4:
226.25 g BaCl22H2O : (2 * 22.99 g Na) + 32.06 g S + (4 * 16.00 g O)
226.25 g BaCl22H2O : 22.99 g Na2SO4
Now we can calculate the amount of BaCl22H2O needed to react with the sulfate in the given samples of sodium sulfate.
Let's assume we have "x" grams of Na2SO4. So, the required amount of BaCl22H2O would be:
x g Na2SO4 : (226.25 g BaCl22H2O)/(22.99 g Na2SO4) * x g Na2SO4
Now, we can calculate the number of moles of BaCl22H2O needed:
(226.25 g BaCl22H2O)/(22.99 g Na2SO4) * x g Na2SO4 * (1 mol BaCl22H2O)/(226.25 g BaCl22H2O)
Lastly, we need to convert the moles of BaCl22H2O to mL of 5% BaCl22H2O solution. We know that the density of the solution is 1.00 g/mL, and the concentration is 5%, which means there are 5 g of BaCl22H2O per 100 mL of solution.
Let's calculate the number of moles of BaCl22H2O needed:
mol BaCl22H2O = (226.25 g BaCl22H2O)/(22.99 g Na2SO4) * x g Na2SO4 * (1 mol BaCl22H2O)/(226.25 g BaCl22H2O)
Finally, let's convert moles of BaCl22H2O to mL of 5% BaCl22H2O solution:
mL BaCl22H2O solution = (mol BaCl22H2O * (100 mL BaCl22H2O solution / 5 g BaCl22H2O)) = (mol BaCl22H2O * 20 mL BaCl22H2O solution/g BaCl22H2O)
For example, if we have 10 grams of sodium sulfate (Na2SO4), we can calculate the required mL of 5% BaCl22H2O solution as follows:
mol BaCl22H2O = (226.25 g BaCl22H2O)/(22.99 g Na2SO4) * 10 g Na2SO4 * (1 mol BaCl22H2O)/(226.25 g BaCl22H2O)
mL BaCl22H2O solution = mol BaCl22H2O * 20 mL BaCl22H2O solution/g BaCl22H2O
To determine the volume of the 5% BaCl2·2H2O solution required to precipitate all the sulfate, we need to understand the chemical reaction that occurs between barium chloride (BaCl2·2H2O) and sodium sulfate (Na2SO4). The reaction equation is as follows:
BaCl2·2H2O + Na2SO4 → BaSO4 + 2NaCl + 2H2O
From the equation, we can see that one mole of barium chloride (BaCl2·2H2O) reacts with one mole of sodium sulfate (Na2SO4) to form one mole of barium sulfate (BaSO4). It is important to note that the 5% concentration of BaCl2·2H2O refers to the mass/volume concentration, i.e., 5 g of BaCl2·2H2O per 100 mL of solution.
To calculate the volume of the 5% BaCl2·2H2O solution required, we need to determine the number of moles of sodium sulfate (Na2SO4) present.
1. Determine the molecular weight of Na2SO4:
- Atomic weight of sodium (Na) = 22.99 g/mol
- Atomic weight of sulfur (S) = 32.06 g/mol
- Atomic weight of oxygen (O) = 16 g/mol
- Molecular weight of Na2SO4 = (2 * 22.99) + 32.06 + (4 * 16) = 142.04 g/mol
2. Calculate the number of moles of Na2SO4:
- Mass of Na2SO4 = Assuming you have a known mass of sodium sulfate
- Moles of Na2SO4 = Mass of Na2SO4 / Molecular weight of Na2SO4
3. Determine the stoichiometry of the reaction:
- From the balanced equation, we know that one mole of BaCl2·2H2O reacts with one mole of Na2SO4.
4. Calculate the number of moles of BaCl2·2H2O required:
- Moles of BaCl2·2H2O = Moles of Na2SO4
5. Determine the mass of BaCl2·2H2O required:
- Mass of BaCl2·2H2O = Moles of BaCl2·2H2O * Molecular weight of BaCl2·2H2O
6. Calculate the volume of the BaCl2·2H2O solution required:
- Volume of BaCl2·2H2O solution (mL) = Mass of BaCl2·2H2O / Density of BaCl2·2H2O solution
Note: Make sure to convert the mass from grams to the same unit as the density (grams to grams or milligrams to milliliters) before dividing by the density.
By following these steps, you can calculate the approximate volume of the 5% BaCl2·2H2O solution required to precipitate all the sulfate.