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.

Answer 1

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.

Answer 2

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.

Answer 3

Yes I found it. Thank you sir for your help.