The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2
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Answer:
1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Explanation:
1) Possible reactions
2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Mass of each metal
a) Mass of Cu
The waste was the unreacted copper.
Mass of Cu = 2.5 g
b) Masses of Al and Fe
We have two relations :
Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g
H₂ from Al + H₂ from Fe = 6.38 L at NTP
i) Calculate the moles of H₂
NTP is 20 °C and 1 atm.
(ii) Solve the relationship
Let x = mass of Al. Then
7.5 - x = mass of Fe
Moles of Al = x/27
Moles of Fe = (7.5 - x)/56
Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18
Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56
∴ x/18 + (7.5 - x)/56 = 0.2652
56x + 18(7.5 - x) = 267.3
56x + 135 - 18x = 267.3
38x = 132.3
x = 3.5 g
Mass of Al = 3.5 g
Mass of Fe = 7.5 g - 3.5 g = 4.0 g
The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g