Answer:
(c) 18.8 g; (a) 0.798; (b) 16 mL
Explanation:
You don't give your experimental data, so I shall assume:
Mass of Al = 1.07 g
20 mL of 3 mol·L⁻¹ KOH
20 mL of 9 mol·L⁻¹ H₂SO₄
The overall equation for the reaction is
Mᵣ: 26.98 474.39
2Al + 2KOH +4H₂SO₄ + 22H₂O ⟶ 2K[Al(SO₄)₂]·12H₂O + 3H₂
m/g: 1.07
(c) Theoretical yield of alum
(i) Moles of Al
[tex]\text{Moles of Al} = \text{1.07 g Al} \times \dfrac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.039 66 mol Al}[/tex]
(ii) Moles of alum
[tex]\text{Moles of alum} = \text{0.039 66 mol Al} \times \dfrac{\text{2 mol alum }}{\text{2 mol Al}} = \text{0.039 66 mol alum \\}[/tex]
(iii) Theoretical yield of alum
[tex]\text{Mass of alum} = \text{0.039 66 mol alum} \times \dfrac{\text{474.39 g alum}}{\text{1 mol alum}} = \textbf{18.8 g alum}[/tex]
(a) Scaling factor for 15.0 g alum
You want a theoretical yield of 15.0 g, so you must scale down the reaction.
[tex]\text{Scale factor} = \dfrac{15.0}{18.8} = \mathbf{0.798}[/tex]
(b) Corrected volumes of NaOH and H₂SO₄
V = 0.798 × 20 mL = 16 mL
