The formula for epsom salt is MgSO4 . 7H20 If 1.250 g of the compound is dissolved in water, calculate the number of milliliters of .200 M Ba(NO3)2 that would be required to precipitate all of the sulfate ions as barium sulfate. I got that right with an answer of 25.4 milliliters.
Then it asks to make the same determination for 1.000 g of alum, which is AlK(SO4)2 .12H2O. I'm not sure .. is the mole to mole ratio 1:2? How do I figure this part out? Any answers would be greatly appreciated.
I get 21.08 which would round to 21.1 mL to three s.f.
so is the answer 1000ml?
or 21.075ml?
what is the chemical equation of this problem
To determine the number of milliliters of 0.200 M Ba(NO3)2 required to precipitate all of the sulfate ions as barium sulfate, you need to consider the stoichiometry of the reaction. First, let's write the balanced chemical equation for the reaction between barium nitrate (Ba(NO3)2) and magnesium sulfate (MgSO4):
Ba(NO3)2 + MgSO4 -> BaSO4 + Mg(NO3)2
From the balanced equation, you can see that 1 mole of Ba(NO3)2 reacts with 1 mole of MgSO4 to form 1 mole of BaSO4. This means the mole ratio between Ba(NO3)2 and MgSO4 is 1:1.
Next, we'll calculate the number of moles of MgSO4 in 1.250 g of the compound.
To do this, we need to know the molar mass of MgSO4:
Molar mass of MgSO4 = (1 mole of Mg) + (1 mole of S) + (4 moles of O)
= 24.31 g/mol + 32.06 g/mol + (4 * 16.00 g/mol)
= 120.37 g/mol
Using the molar mass, we can calculate the number of moles of MgSO4:
Moles of MgSO4 = (Mass of MgSO4) / (Molar mass of MgSO4)
= 1.250 g / 120.37 g/mol
≈ 0.01037 moles
Since the mole ratio between Ba(NO3)2 and MgSO4 is 1:1, we need an equal number of moles of Ba(NO3)2 to react with the MgSO4. Therefore, we need 0.01037 moles of Ba(NO3)2.
Now, we can calculate the volume of 0.200 M Ba(NO3)2 needed to obtain 0.01037 moles. The equation we'll use is:
Moles of solute = Molarity × Volume (in liters)
Rearranging the equation to solve for volume, we get:
Volume (in liters) = Moles of solute / Molarity
Volume (in milliliters) = (Moles of solute / Molarity) × 1000
Plugging in the values, we have:
Volume (in milliliters) = (0.01037 moles / 0.200 moles/L) × 1000
≈ 51.85 milliliters
Therefore, you would need approximately 51.85 milliliters of 0.200 M Ba(NO3)2 to precipitate all of the sulfate ions as barium sulfate when 1.250 g of MgSO4 is dissolved in water.
Now, let's move on to the second part of the question, involving 1.000 g of alum (AlK(SO4)2 .12H2O).
For the second part, you correctly identified that the mole-to-mole ratio between aluminum potassium sulfate (alum) and sulfate ions is 1:2.
The balanced chemical equation for the reaction between barium nitrate and alum can be written as:
Ba(NO3)2 + AlK(SO4)2 -> BaSO4 + Al(NO3)3 + KNO3
From the balanced equation, you can see that 1 mole of Ba(NO3)2 reacts with 1 mole of AlK(SO4)2 to form 1 mole of BaSO4.
Since there is a 1:2 mole ratio between AlK(SO4)2 and sulfate ions, you need to multiply the moles of AlK(SO4)2 by 2 to obtain the moles of sulfate ions.
Following the same steps as before, calculate the moles of AlK(SO4)2 using its molar mass:
Molar mass of AlK(SO4)2 = (1 mole of Al) + (1 mole of K) + (2 moles of S) + (8 moles of O)
= 26.98 g/mol + 39.10 g/mol + (2 * 32.06 g/mol) + (8 * 16.00 g/mol)
= 258.20 g/mol
Moles of AlK(SO4)2 = (Mass of AlK(SO4)2) / (Molar mass of AlK(SO4)2)
= 1.000 g / 258.20 g/mol
≈ 0.00387 moles
Since the mole-to-mole ratio between AlK(SO4)2 and sulfate ions is 1:2, the number of moles of sulfate ions is twice the moles of AlK(SO4)2:
Moles of sulfate ions = 2 * 0.00387 moles
≈ 0.00774 moles
Now, you can proceed to calculate the volume of 0.200 M Ba(NO3)2 required to precipitate all of the sulfate ions. Using the same equation as before:
Volume (in milliliters) = (Moles of solute / Molarity) × 1000
Plugging in the values, we get:
Volume (in milliliters) = (0.00774 moles / 0.200 moles/L) × 1000
≈ 38.7 milliliters
Therefore, you would need approximately 38.7 milliliters of 0.200 M Ba(NO3)2 to precipitate all of the sulfate ions as barium sulfate when 1.000 g of alum (AlK(SO4)2 .12H2O) is dissolved in water.
Remember, these calculations assume complete and ideal reaction conditions.
1:2 or 2:1 depending upon which way you do it.
moles alum = 1.000/molar mass alum.
moles BaSO4 = 2 x moles alum
M Ba(NO3)2 = moles Ba(NO3)2/L Ba(NO3)2
Solve for L and convert to mL.