Question

5) Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 43.0 mL of hydrogen gas over water at 27°C and 751 mmHg. How many grams of aluminum reacted? The partial pressure of water at 27°C is 26.8 mmHg.

__gAl

Thank you!

Answer 1

We know that
PV = nRT
n = PV/RT

So, moles of H2(g) is

P = 751 mmHg = (751 / 760) atm = 0.988 atm
V = 43 mL = 0.043 L
R = 0.0821 L atm mol^-1 K^-1
T = 27 ^oC = (27 + 273) K = 300 K

So,
n = [ (0.988 atm) x (0.043 L) ] / [ (0.0821 L atm mol^-1 K^-1) x (300 K) ]
n = 0.0017 moles of H2

Now, 2 Al + 6 HCl ----> 2 AlCl3 + 3 H2(g)
In the above balance reaction equation:
3 moles H2 is produced from 2 moles Al.
1 moles H2 is produced from (2/3) moles Al.
0.0017 moles H2 is produced from (0.0017 x 2) / 3 moles Al.
0.0017 moles H2 is produced from 0.00113 moles Al.

Finally,
0.00113 moles Al x (27.0 g/mole) = 0.03051 g Al