What volume (in mL) of 0.150 M HCl would be required to completely react with 5.45 g of Al in the following chemical reaction?

Question

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

Answer 1

4040 mL (4.04 L)

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

Step 1 — moles of aluminium

n(Al) = 5.45 g / 26.98 g/mol = 0.202 mol

Step 2 — mole ratio

The coefficients give 6 HCl per 2 Al, which reduces to 3:1:

n(HCl) = 0.202 mol × (6 mol HCl / 2 mol Al) = 0.606 mol

Step 3 — moles to volume

Molarity is mol/L, so dividing moles by molarity returns litres:

V = 0.606 mol / 0.150 mol/L = 4.04 L = 4040 mL

The question asks for millilitres, so finish the conversion — leaving the answer as 4.04 is a common way to lose the mark on an otherwise correct problem.

Does the size make sense?

Yes, and it is worth pausing on. Four litres of acid to dissolve about 5 grams of metal sounds excessive until you notice the acid is dilute — 0.150 M is fairly weak, so it takes a lot of solution to deliver 0.6 mol of HCl. Concentrated hydrochloric acid is around 12 M, and at that strength you would need only about 50 mL.

The direction to watch

HCl must come out larger in moles than Al, because the equation consumes three acid molecules per metal atom. If your HCl figure had come out smaller than 0.202 mol, the ratio was inverted — that would give 0.0673 mol and 449 mL, off by a factor of nine.

The general shape: grams → moles → (mole ratio) → moles → volume. The balanced equation only operates in the middle step, and molarity is what converts moles to volume at the end — exactly the reverse of a problem that hands you an acid volume and asks for the mass of metal.