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
To determine the volume of 5% BaCl2·2H2O solution required to precipitate all the sulfate, we need to calculate the stoichiometry of the reaction between barium chloride (BaCl2) and sodium sulfate (Na2SO4).
The balanced equation for the reaction between BaCl2 and Na2SO4 can be written as follows:
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
From the equation, we can see that one mole of BaCl2 reacts with one mole of Na2SO4 to produce one mole of BaSO4. Therefore, the stoichiometric ratio between BaCl2 and BaSO4 is 1:1.
First, let's calculate the molar mass of BaCl2·2H2O, which is 244.25 g/mol.
Next, we need to determine the number of moles of BaSO4 formed. To do this, we'll use the mass of sodium sulfate.
Assuming you have a specific mass of sodium sulfate, take the mass of Na2SO4 in grams and divide it by the molar mass of Na2SO4.
Once you have the moles of Na2SO4, you can determine the moles of BaCl2 required, since the stoichiometric ratio is 1:1.
Now we need to convert the moles of BaCl2 into volume using the density of the BaCl2 solution.
Finally, to convert the volume into milliliters (mL), multiply the volume by 1000.
Keep in mind that this calculation assumes that the reaction goes to completion and that there are no other factors influencing the precipitation.