What volume of 3.0M Na2SO4 must be added to 25ml of 1.0M BaCl2 to produce 5.0g of BaSO4?
The answer is 7.1ml but I don't get it very well.
I thought I could use mole ratio of the balanced equation, but it did not work.
Thanks a lot!
To solve this problem, you are correct that you need to use the mole ratio from the balanced equation. However, the balanced equation is not given here, so let's start by writing it.
The balanced equation for the reaction between sodium sulfate (Na2SO4) and barium chloride (BaCl2) to produce barium sulfate (BaSO4) can be written as:
Na2SO4 + BaCl2 -> BaSO4 + 2NaCl
From the balanced equation, we can see that one mole of sodium sulfate reacts with one mole of barium chloride to produce one mole of barium sulfate.
Now let's calculate the amount of barium sulfate that will be produced. We are given that the mass of barium sulfate produced is 5.0g. To calculate the number of moles of BaSO4, we need to use its molar mass.
The molar mass of BaSO4 = 137.33 g/mol (barium) + 32.06 g/mol (sulfur) + 4 * 16.00 g/mol (oxygen) = 233.38 g/mol
Number of moles of BaSO4 = mass / molar mass = 5.0g / 233.38 g/mol ≈ 0.0214 mol
Since the mole ratio of BaSO4 to Na2SO4 is 1:1, the amount of Na2SO4 required will also be 0.0214 mol.
Now let's calculate the volume of 3.0M Na2SO4 that contains 0.0214 mol. We can use the equation:
moles = concentration * volume
0.0214 mol = 3.0M * volume
volume = 0.0214 mol / 3.0M ≈ 0.0071 L = 7.1 mL
Therefore, approximately 7.1 mL of 3.0M Na2SO4 must be added to 25 mL of 1.0M BaCl2 to produce 5.0g of BaSO4.
To solve this problem, we need to use the concept of stoichiometry. The balanced equation for the reaction between BaCl2 and Na2SO4 is:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
Based on the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4.
Step 1: Calculate the number of moles of BaSO4 required.
The molar mass of BaSO4 is:
Ba: 1 * 137.33 g/mol = 137.33 g/mol
S: 1 * 32.06 g/mol = 32.06 g/mol
O: 4 * 16.00 g/mol = 64.00 g/mol
Total molar mass of BaSO4 = 233.39 g/mol
Given that we need 5.0 g of BaSO4, we can calculate the number of moles:
moles of BaSO4 = mass / molar mass = 5.0 g / 233.39 g/mol = 0.0214 mol
Step 2: Use the mole ratio to find the number of moles of Na2SO4 needed.
From the balanced equation, we know that 1 mole of BaSO4 is produced from 1 mole of Na2SO4. Therefore, the number of moles of Na2SO4 required is also 0.0214 mol.
Step 3: Calculate the volume of 3.0M Na2SO4 needed to provide 0.0214 mol.
The concentration of Na2SO4 is given as 3.0M, meaning there are 3.0 moles of Na2SO4 dissolved in 1 liter of solution.
moles = concentration * volume
0.0214 mol = 3.0 mol/L * volume
volume = 0.0214 mol / 3.0 mol/L = 0.0071 L = 7.1 mL
Therefore, you would need to add 7.1 mL of 3.0M Na2SO4 to the solution to produce 5.0 g of BaSO4.
Note that one mole of the sulfate ion reacts with one mole of the barium ion.
Number of moles of product (= number of moles of barium ion) =
5.0 g/ (RMM of barium sulfate)
5.0/233.40 = 0.0214 moles
[note that this is less than
25 x 1.0/1000, the number of moles of barium ions present]
number of moles of sodium sulfate (= number of moles of sulfate ion)is
M x V/1000
3.0 x V/1000 = 0.0214
so V= 7.1 ml to two sig figs.