Question

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. What mass of Al(s) is required to produce 643.0 mL of H2(g) at STP? I have the equation as

2AL + 6Hcl = 2AlCl3 + 3H2

Answer 1

we know that

for gases

PV = nRT

given

volume = 643 ml = 0.643 L

now at

STP conditions

so pressure (P ) = 1 atm

temperature (T) = 273.15 K

so

1 x 0.643 = n x 0.0821 x 273.15

n = 0.02867

so

moles of H2 gas produced = 0.02867

now

consider the given reaction

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

we can see that

moles of H2 produced = 1.5 x moles of Al taken

0.02867 = 1.5 x moles of Al taken

moles of Al taken = 0.019115

now

mass = moles x molar mass

molar mass of Al = 27 g/mol

so

mass of Al = 0.019115 x 27

mass of Al = 0.516 g

so

0.516 grams of Al is required