Assuming the sample is pure sodium sulfate, how many mL of 5% BaCl2.2H20 solution would be required to precipitate the sulfate. Density of barium shloride solution of 1.00 g/mL.
That all depends upon how much of the Na2SO4 is involved. moles Na2SO4 = grams/molar mass
Determine the M of the BaCl2.2H2O soln. 5% means you have 5.0 g BaCl2.2H2O/100 mL. moles = grams/molar mass; then M BaCl2.2H2O = moles BaCl2.2H2O/0.1L = ?M.
Write the equation and convert moles Na2SO4 to moles BaCl2.2H2O. Then
M BaCl2.2H2O = moles Ba^2+ required/L soln and solve for L.
To determine how many mL of the 5% BaCl2.2H2O solution would be required to precipitate the sulfate, we need to use stoichiometry and the equation:
BaCl2 + Na2SO4 --> BaSO4 + 2NaCl
We can start by calculating the number of moles in the sodium sulfate (Na2SO4) sample. Let's assume we have a known mass of sodium sulfate sample (m).
1. Determine the molar mass of sodium sulfate (Na2SO4). The molar mass can be calculated by adding the atomic masses of each element:
Na (sodium) = 22.99 g/mol
S (sulfur) = 32.07 g/mol
O (oxygen) = 16.00 g/mol
Molar mass of Na2SO4 = (2 x 22.99) + 32.07 + (4 x 16.00) = 142.04 g/mol
2. Calculate the number of moles of sodium sulfate (Na2SO4) using the given mass (m) and the molar mass:
moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
Next, we need to determine the stoichiometry of the reaction. From the balanced equation, we see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to form 1 mole of BaSO4.
3. Since the reaction is 1:1, the number of moles of BaCl2 required is the same as the number of moles of Na2SO4.
4. Now, we need to calculate the mass of BaCl2 required using the molar mass of BaCl2.
molar mass of BaCl2 = 137.33 g/mol
mass of BaCl2 = moles of BaCl2 * molar mass of BaCl2
5. Calculate the volume of BaCl2.2H2O solution needed based on its density and the mass of BaCl2 required.
volume of BaCl2.2H2O = mass of BaCl2 / density of BaCl2.2H2O
Keep in mind that the given BaCl2.2H2O solution has a concentration of 5%, which means it contains 5 g of BaCl2 in every 100 g of solution.
6. Finally, convert the volume to mL if necessary.
Note: This calculation assumes that the reaction goes to completion and that there are no side reactions or losses during the precipitation process.
To determine how many mL of 5% BaCl2.2H20 solution would be required to precipitate the sulfate, we need to consider the reaction between sodium sulfate (Na2SO4) and barium chloride (BaCl2). The balanced chemical equation for this reaction is:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
This equation shows that one mole of sodium sulfate reacts with one mole of barium chloride to produce one mole of barium sulfate and two moles of sodium chloride.
To calculate the volume of the BaCl2.2H20 solution needed, we follow these steps:
1. Determine the molar mass of Na2SO4:
- Na (sodium) = 22.99 g/mol
- S (sulfur) = 32.07 g/mol
- O (oxygen) = 16.00 g/mol (x 4) = 64.00 g/mol
- Total molar mass = 22.99 g/mol + 2(32.07 g/mol) + 64.00 g/mol = 142.06 g/mol
2. Calculate the moles of Na2SO4:
- Given that the sample is pure sodium sulfate, let's assume we have "x" moles.
3. From the balanced equation, we can see that the ratio between Na2SO4 and BaCl2 is 1:1, meaning that the moles of BaCl2 needed to react fully will also be "x".
4. Determine the molar mass of BaCl2.2H20:
- Ba (barium) = 137.33 g/mol
- Cl (chloride) = 35.45 g/mol (x 2) = 70.90 g/mol
- H (hydrogen) = 1.01 g/mol (x 2) = 2.02 g/mol
- O (oxygen) = 16.00 g/mol
- Total molar mass = 137.33 g/mol + 70.90 g/mol + 2.02 g/mol + 16.00 g/mol = 226.25 g/mol
5. Calculate the mass of BaCl2.2H20 needed:
- Given that BaCl2.2H20 is present in a 5% solution, this means there are 5g of BaCl2.2H20 in every 100 mL of the solution.
- The density of the solution is given as 1.00 g/mL, so the mass of BaCl2.2H20 in 100 mL is 100g.
- Therefore, we have 5g of BaCl2.2H20 in 100g of the solution.
- Using the equation: mass = (percentage/100) x total mass
mass of BaCl2.2H20 = (5/100) x 100g = 5g
6. Calculate the moles of BaCl2.2H20:
- Using the molar mass of BaCl2.2H20, we find:
moles of BaCl2.2H20 = mass of BaCl2.2H20 / molar mass
= 5g / 226.25 g/mol
= 0.0221 moles
7. Since the ratio between Na2SO4 and BaCl2 is 1:1, we know that:
- 1 mole of Na2SO4 reacts with 1 mole of BaCl2
- 0.0221 moles of Na2SO4 will require 0.0221 moles of BaCl2
8. Convert the moles of BaCl2 to volume using the density given:
- volume (in mL) = moles of BaCl2 x molar mass of BaCl2 / density of the solution
= 0.0221 moles x 226.25 g/mol / 1.00 g/mL
= 4.99 mL
Therefore, approximately 4.99 mL of the 5% BaCl2.2H20 solution would be required to precipitate the sulfate in the sample.