Some help with Gas Equation

Give: 2 Al(s) + 6 HCL(aq) --> 3 H2(aq) + 2 ALCL3(aq). If 2.00 grams of AL(s) react with excess HCL(aq), how many liters of H2 gas can bw collected at 27.0 degrees C and 1.00 atm pressure? (molar masses, (g/mol): Al, 26.98, HCL 36.46, H2 2.02, AlCL3 133.33) how do u do this? work would be great!

first H2(g) is the product

2.00 g Al = 2.00/26.98 mole

2 mole Al produces 3 mole H2

moles H2 produced = 3*2.00/(26.98*2) = 0.11 mole

PV=nRT ........ R=0.08205746 L atm K−1 mol−1 .... T=273+27=300 K
1*V=0.11*0.08205746*300

V = 2.7 L

The no of moles of H2 gas produced by the reaction is 3/2 times the moles of Al consumed. The no of moles of Al is 2/(at wt of Al) =2/26.98 = 0.07413 mol; 3/2 times this is 0.1112 mol of H2 produced, This is "n" in the gas formula P*V = n*R*T

V = n*R*T/P

n = 0.1112
T = 300.15 ºK
P = 1.00 atm
R = 0.08206 L*atm/ºK*mol

V = 2.74 L

Balanced equation:
2 Al + 6 HCl = 2 AlCl3 + 3 H2
2mol Al produce 3 mol H2

Molar mass Al = 26.98g/mol
2.0g Al = 2/26.98 = 0.074 mol Al
This will produce 0.074 * 3/2 = 0.111 mol H2

Calculate volume using the gas law equation:
PV = nRT
1.0*V = 0.111*0.082057*300
V = 2.74 litres

Volume of H2 produced = 2.74L