Chemistry Help (Molarity)?

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(0.0750 L) x (0.54 mol/L HCl) = 0.0405 mol HCl

0.0405 mole of HCl would react completely with 0.0405 x (2/6) = 0.0135 mol Al, but there is more Al present than that, so Al is in excess and HCl is the limiting reactant. [You guessed wrong about the limiting reactant.]

a.
(0.0405 mol HCl) x (3 mol H2 / 6 mol HCl) x (22.414 L/mol) = 0.45 L H2

b. Supposing the AlCl3 ionizes completely: 2 AlCl3 → 2 Al3 6 Cl-
(0.0405 mol HCl) x (2 mol Al3 / 6 mol HCl) / (0.0750 L) = 0.18 mol/L Al3

c. HCl → H Cl-
The concentration of the Cl- ions is the same as the original concentration of the HCl since the volume didn't change, so 0.54 M Cl-.

(a)
2Al + 6HCl → 3H₂(g) + 2AlCl₃

Molar mass of Al = 26.98 g/mol
Initial number of moles of Al = (10.0 g) / (26.98 g/mol) = 0.371 mol
Initial number of moles of HCl = (0.54 mol/L) × (75.0/1000 L) = 0.0405 mol

According to the equation, mole ratio Al : HCl = 2 : 6 = 1 : 3
For complete reaction of HCl :
Number of moles of Al needed = 0.0405 × 3 = 0.1215 mol < 0.371 mol
Hence, Al is in excess, and thus HCl is the limiting reactant (completely reacts).

According to the equation, mole ratio HCl : H₂ = 6 : 3 = 2 : 1
Number of moles of HCl reacted = 0.0405 mol
Number of moles of H₂ produced = (0.0405 mol) × (1/2) = 0.02025 mol

1 mole of H₂ gas occupies 22.4 L of volume at STP.
Volume of H₂ produced = (0.02025 mol) × (22.4 L/mol) = 0.454 L


(b)
According to the equation, mole ratio HCl : AlCl₃ = 6 : 2 = 3 : 1
Besides, 1 mole of AlCl₃ dissociates to give 1 mole of Al³⁺.
Hence, mole ratio HCl : Al³⁺ = 3 : 1

Number of moles of HCl reacted = 0.0405 mol
Number of moles of Al³⁺ = (0.0405 mol) × (1/3) = 0.0135 mol

Volume of the solution = 75 mL = 0.075 L
Molarity of Al³⁺ = (0.0135 mol)/(0.075 L) = 0.18 M


(c)
Cl⁻ is the spectator ion, which does not react at all.
It concentration is unchanged.
Hence, molarity of Cl⁻ = original molarity of HCl = 0.54 M

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