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EquilibriumFRQ

Answer the following questions about Fe and Al compounds. Use the following reactions that involve FEquilibrium Chemistry Question

Problem Context

Answer the following questions about Fe and Al compounds.

Use the following reactions that involve Fe and Al compounds to answer parts (b) and (c).

In distilled water
Reaction 1: Fe2O3(s) + 3 H2O(l) → 2 Fe(OH)3(s)
Reaction 2: Al2O3(s) + 3 H2O(l) → 2 Al(OH)3(s)

In base
Reaction 3: Fe(OH)3(s) + NaOH(aq) → no reaction
Reaction 4: Al(OH)3(s) + NaOH(aq) → NaAl(OH)4(aq)

In acid
Reaction 5: Fe(OH)3(s) + 3 HCl(aq) → FeCl3(aq) + 3 H2O(l)
Reaction 6: Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l)
Reaction 7: NaAl(OH)4(aq) + HCl(aq) → Al(OH)3(s) + NaCl(aq) + H2O(l)

When heated
Reaction 8: 2 Fe(OH)3(s) -heat→ Fe2O3(s) + 3 H2O(g)
Reaction 9: 2 Al(OH)3(s) -heat→ Al2O3(s) + 3 H2O(g)

a.2 pts

Model Answer

Fe3+ and Al3+ have similar sizes (radii).
Fe3+ and Al3+ have the same charge.

1 point is earned for each reason.

b(i,ii).2 pts

Model Answer

Al3+ will be present in higher concentration.
Al(OH)3 has the same stoichiometry as Fe(OH)3 but a greater Ksp.

1 point is earned for the correct choice and explanation. || Al(OH)3(s) → Al3+(aq) + 3 OH−(aq)
OR
Al2O3(s) + 3 H2O(l) → 2 Al3+(aq) + 6 OH−(aq)

1 point is earned for a balanced equation.

c(i).1 pt

Model Answer

This approach only works when there is a significant difference in water solubility between two substances.

1 point is earned for a correct explanation.

c(ii).3 pts

Model Answer

Step 3: Add HCl to the filtrate until a precipitate of Al(OH)3 forms. (Reaction 7)
Step 4: Filter out the solid Al(OH)3. (Reaction -)
Step 5: Heat the solid Al(OH)3 to form Al2O3. (Reaction 9)

1 point is earned for each correct row (description plus reaction number, if applicable).

c(iii).2 pts

Model Answer

5.5 g Al2O3 × (1 mol Al2O3 / 101.96 g Al2O3) × (2 mol Al / 1 mol Al2O3) × (26.98 g Al / 1 mol Al) = 2.9 g Al
(2.9 g Al / 10.0 g mixture) × 100 = 29%

1 point is earned for the number of grams of Al.
1 point is earned for the mass percent Al.

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