Suppose we collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

Question

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express
your answer in liters.

Answer 1

2Al + 6HCl => 2AlCl3 + 3H2
(1.35/27)mole Al = 0.05molAl => 3/2(.05)mol H2 = 0.075mol H2
Vol H2 = nRT/P = [(0.075)(0.08206)(294)(760)]/(743) = 1.85L H2

Answer 2

1.9L