How many liters of h2 gas at stp can be made if 2.698 g Al is dissolved in HCl solution

how many liters of h2 gas at stp can be made if 2.698 g Al is dissolved in HCl solution?

I'm a bit stuck. :/

How many moles of Al do we have? Number of moles = mass / mass of 1 mole = 2.698 / 26.98 = 0.1 mole

Al + HCl → AlCl3 + H2
Balance the equation
2 Al + 6 HCl → 2 AlCl3 + 3 H2
According to the balanced equation above, 2 moles of Al will react with 6 moles of HCl to produce 2 moles of AlCl3 and 3 moles of H2.

Mole ratio of Al to H2 = 2 : 3
2 moles of Al will produce 3 moles of H2, so 0.1 mole of Al will produce x moles of H2.
Set up a proportion: 2 / 3 = 0.1 / x, x = 3 * 0.1 ÷ 2 = 0.15

0.15 mole of H2 is produced at STP.
The volume of 1 mole of a gas at STP = 22.4 Liters
The volume of 0.15 mole of H2 at STP = 0.15 * 22.4

2 Al + 6 HCl -----> 2AlCl3 + 3 H2
moles Al: 2.698 g/26.98 g/mole= 0.1
from bal. rxn., 3 moles of H2 are produced per 2 moles Al (3:2)
0.1 moles Al x 3/2 =0.15 moles H2 are produced
@ STP conditions 1 mole occupies 22.4 L, therefore 0.15 moles H2 x 22.4 l/Mole= 3.36 L

STP is 273 K and 1 atm.

2Al + 6HCl --> 3H2 + 2AlCl3

2.698 grams Al * (mol Al/27 grams) * (3 moles H2/2 moles Al) = 0.15 moles H2

pv=nRT
(1 atm)(v) = (0.15 moles)(.0821 atm*L/mol*K)(273K)

v = 3.36 Liters