If 0.755 g of Al3+ dissolved in 960 mL of water reacts stoichiometrically according to the balanced equation, what volume (mL) of 0.0875 M aqueous Cl- is required?
Al2(SO4)3(aq) + 3 BaCl2(aq) → 2 AlCl3(aq) + 3 BaSO4(s)
It's late for me and easy to get confused. Check me out thoroughly.
0.755 g Al^+/molar mass Al x [1 mol Al2(SO4)3/2 mols Al^+3] = moles Al2(SO4)3.
That x 3/2 = moles BaCl2 which contains twice that much Cl^-.
Then moles Cl^-/L = M. Solve for L and multiply by 1000. Check my thinking. I'm off to bed. It's approaching 1:00 AM here.
To find the volume of 0.0875 M aqueous Cl- required, we first need to determine the stoichiometry of the reaction between Al3+ and Cl-. From the balanced equation:
Al2(SO4)3(aq) + 3 BaCl2(aq) → 2 AlCl3(aq) + 3 BaSO4(s)
we can see that for every 2 moles of AlCl3 produced, 6 moles of Cl- ions are consumed.
Next, we need to calculate the moles of Al3+ dissolved in 960 mL of water. To do this, we need to convert the mass of Al3+ to moles using its molar mass.
Given:
Mass of Al3+ = 0.755 g
Molar mass of Al3+ = 26.98 g/mol
moles of Al3+ = Mass / Molar mass = 0.755 g / 26.98 g/mol ≈ 0.028 mol
Since the balanced equation shows that 2 moles of AlCl3 are produced for every 1 mole of Al3+, we can infer that 0.014 moles of AlCl3 will be produced.
Now, we know that for every 2 moles of AlCl3, 6 moles of Cl- ions are consumed. Therefore, for 0.014 moles of AlCl3, the amount of Cl- ions consumed would be:
moles of Cl- = moles of AlCl3 × (6 moles Cl- / 2 moles AlCl3) = 0.014 mol × (6 mol Cl- / 2 mol AlCl3) = 0.042 mol
Finally, we can calculate the required volume of 0.0875 M aqueous Cl-. The concentration (Molarity) of a solution is defined as moles per liter (mol/L). So, we need to determine the volume of the solution containing 0.042 moles of Cl- ions when the concentration is 0.0875 M.
Moles = Molarity × Volume (in liters)
0.042 mol = 0.0875 mol/L × Volume (in L)
Volume (in L) = 0.042 mol / 0.0875 mol/L ≈ 0.48 L
Lastly, we need to convert the volume from liters to milliliters:
Volume (in mL) = Volume (in L) × 1000 mL/L = 0.48 L × 1000 mL/L ≈ 480 mL
Therefore, approximately 480 mL of 0.0875 M aqueous Cl- is required.