2Al + 2OH- + 2H2O → 3H2 + 2AlO2- How many mL of H2 are released at 750.0mmHg and 50.0oC when Drano, with 0.150g of Al, is used for cleaning?

Answer 1

2Al + 2OH^- + 2H₂O ==> 3H₂ + 2AlO₂^-

0.150 g Al x 1 mol/26.98 g = 0.0055596 moles Al

0.0055596 moles Al x 3 mol H2/2 mol Al = 0.0083395 moles H2

PV = nRT

T = 50 + 273 = 323K

P = 750 mm/760 mm/atm = 0.9869 atm

V = nRT/P = (0.0083395)(0.0821)(323)/0.9868 = 0.224 L = 224 ml