According to the reaction represented above, about how many grams of aluminum (atomic mass 27 g) are necessary to produce 0.50 mol of hydrogen gas at 25oC and 1.0 atm?

(A) 1.0 g

(B) 9.0 g

(C) 14 g

(D) 27 g

(E) 56 g

9 grams are necessary to produce 0.50 mol of H₂

Explanation:

I guess, this can be a certain reaction to that:

6 HCl + 2Al → 2AlCl₃ + 3H₂

Let's think a rule of three:

If 3 moles of H₂ are produced by 2 moles of aluminum

0.5 moles of H₂ are produced by (0.5. 2) / 3 = 0.333 moles

Now that we have the moles, let's know the grams by this.

Molar mass. moles = grams

27 g/m. 0.3333 m = 9 g