A piece of Al weighing 2.7g is titrated with 75 mL of H2SO4 (sp.gr. 1.18 g/mL and 24.7% H2SO4 by mass). After the metal is completely dissolved the solution is dilute to 400mL. Calculate molarity of free H2SO4 in solution.
2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2
mols Al = grams/atomic mass = 0.1
mols H2SO4 used = 0.1 mol Al x (3 mols H2SO4/2 mols Al) = 0.15 mol H2SO4 used.
mols H2SO4 initially = 1.18 g/mL x 75 mL x 0.247 x (1/molar mass H2SO4) = approx 0.2 but you need a better answer than that.
mols H2SO4 not used = approx 0.2-0.15 = 0.05 (remember to redo these calculations)
Then you have 0.05 mol/0.400 L = ? M.
By the way, I posted a response to your NaCl/CaCl2 ratio problem. If you haven't found it let me know and I can give youa link.
Yes I found it. Thank you sir for your help.
To calculate the molarity of free H2SO4 in the solution, we need to perform a series of steps.
Step 1: Calculate the moles of Al used in the reaction.
To do this, we can use the molar mass of Al, which is 26.98 g/mol.
moles of Al = mass of Al / molar mass of Al
= 2.7 g / 26.98 g/mol
≈ 0.1001 mol
Step 2: Calculate the moles of H2SO4 used in the reaction.
In order to determine the moles of H2SO4, we must first calculate the mass of the H2SO4 solution used.
mass of H2SO4 solution = volume of H2SO4 solution × density of H2SO4 solution
= 75 mL × 1.18 g/mL
= 88.5 g
Next, we can calculate the mass of H2SO4 in the solution, considering it is 24.7% H2SO4 by mass.
mass of H2SO4 = (24.7 / 100) × mass of H2SO4 solution
= (24.7 / 100) × 88.5 g
= 21.8695 g
Now, we can calculate the moles of H2SO4.
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
= 21.8695 g / (2 × 1 + 32.06 + 4 × 16.00) g/mol
≈ 0.1813 mol
Step 3: Calculate the molarity of H2SO4 in the solution.
Since the solution is diluted to 400 mL (0.4 L), we can use the formula for molarity:
Molarity = moles of solute / volume of solution in liters
Molarity of H2SO4 = moles of H2SO4 / volume of solution
= 0.1813 mol / 0.4 L
≈ 0.4533 M
Therefore, the molarity of free H2SO4 in the solution is approximately 0.4533 M.