Question

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced.

2Al(s)+6HCl(aq)--->2AlCl3(aq)+3H2

What mass of Al(s) is required to produce 590.0 mL of H2(g) at STP?

? gAl

Answer 1

From STP conditions, w know that

1 mol of gas = 22.4L at STP

so..

590 mL = 0.590 L

relate to mol of H2

1 mol / 22.4 Liters * 0.59 Liters = 0.026339285 moles of H2

so...

the reaction states that

2 mol of Al --> 3 mol of H2

so..

2/3 mol of Al --> 1 mol of H2

so --> 0.026339285 mol of H2 --> 2/3*0.026339285 mol of Al = 0.017559 mol of Al

mass = mol*MW

Mw Al = 26.981539 g/mol

so

mass = 0.017559*26.981539 = 0.4737 g of Aluminium required for 590 mL of H2 at STP