Answer:
A. 2 Al (s) + 6HCl (aq) → 2AlCl₃ (s) ↓ + 3H₂ (g)
B. Al is the excess reactant and HCl is the limiting.
C. 0.672 L of H₂ produced at STP
D. 2.67 g of AlCl₃ are made in this reaction.
E. 4.86 g of Al remain after the reaction goes complete.
Explanation:
We star from the reaction:
2 Al (s) + 6HCl (aq) → 2AlCl₃ (s) ↓ + 3H₂ (g)
2 moles of aluminum, react with 6 moles of HCl in order to produce 2 moles of aluminum chloride and 3 mol of H₂ gas.
We determine moles of each reactant:
[HCl] = 0.2M → 0.2 mol/L . 0.3L = 0.060 moles
(we converted 300 mL to 0.3L)
5.4 g of Al . 1mol / 26.98g = 0.200 moles
Ratio is 2:6 (3). 2 mol of Al react to 6 mol of HCl
0.2 moles of Al may react with (0.2 . 6) /2 = 0.6 mol of acid
We have 0.06 moles, and we need 0.6 mol of acid, so the HCl is the limiting reactant. Then, the Al is the excess:
6 moles of HCl need 2 moles of Al to react
Then 0.06 moles of HCl will react to (0.06 . 2) /6 = 0.02 moles
If we have 0.2 moles of Al, and we need 0.02 moles for the reaction, then
(0.2 - 0.02) = 0.18 moles remain after the reaction is complete.
0.18 mol . 26.98g /1mol = 4.86 g of Al remain after the reaction goes complete.
As the limting reactant is the HCl, we work with it to determine the mass of salt which is produced:
6 mol of HCl can produce 2 mol of chloride
Then 0.06 moles of HCl will produce (0.06 . 2) /6 = 0.02 mol of AlCl₃
We convert to mass: 0.02 mol . 133.33g/1mol = 2.67 g of AlCl₃ are made in this reaction.
Let's find out the volume of hydrogen produced, at STP
6 moles of HCl can produce 3 moles of H₂
0.06 moles of HCl will produce (0.06 . 3) /6 = 0.03 moles of H₂
1 mol of any gas at STP occupies 22.4L
0.03 moles of H₂ will ocuppy (22.4 L . 0.03 mol)/1mol = 0.672L
