Step 1 :
The balanced chemical equation for the above problem can be written as:
2Al + 6HCl → 2AlCl₃ + 3H₂
Step 2 :
Since 15 grams of Al metal react with the excess of HCl, therefore Al will be the limiting reactant.
The moles of Al that reacted will be = 15/27
=0.55 moles of Al.
Step 3:
Molar ratio of Al and H₂ will be 2:3.
Step 4:
Therefore moiles of H2 produced by 0.55 moles of Al
0.55 moles of Al × (3 moles of H₂ ÷2 moles of Al)
=0.55×1.5
=0.825 moles of H₂.
Step 5:
By using the ideal gas equation we can obtain the volume of H2 produced.
pV=nRT (Ideal Gas Equation)
V=(nRT) ÷ p
Step 6:
n=0.825 moles
R=0.0821 litre atm mol⁻¹ K⁻¹
T=27 degree celsius = 27+273 K= 300 K.
p = 680 torr × ( 1 atm ÷ 760 torr )
=0.894 atm.
Step 7:
V=[0.0825 moles × 0.0821 atm L mol⁻¹ K⁻¹×300 K] ÷ 0.894 atm
V=2.27×10⁻³ litres.
