What volume (in mL) of 0.250 M HCl would be required to completely react with 4.90 g of Al in the following chemical reaction?

Question

2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)

Answer 1

The 4.90g Al * (1 mol Al / 26.98g Al) * (6 mol HCl / 2 mol Al) requires 0.545 moles of HCl. With molarity 0.250 moles/liter, volume = 0.545 mol / 0.250 mol/L = 2.18 L = 2,180 mL.