Question

Scaled Synthesis of Alum. Show your calculations for:a.the experimental scaling factor giving rise to a 15.0 g theoretical yield;b.the corrected volumes of KOH and H2SO4; andc.the theoretical yield of alum based on the actual amount of Al used.Make sure you carefully show each step for these calculations.

Answer 1

(c) 18.8 g; (a) 0.798; (b) 16 mL

Step-by-step 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

`\text{Moles of Al} = \text{1.07 g Al} * \frac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.039 66 mol Al}`

(ii) Moles of alum

`\text{Moles of alum} = \text{0.039 66 mol Al} * \frac{\text{2 mol alum }}{\text{2 mol Al}} = \text{0.039 66 mol alum \\}`

(iii) Theoretical yield of alum

`\text{Mass of alum} = \text{0.039 66 mol alum} * \frac{\text{474.39 g alum}}{\text{1 mol alum}} = \textbf{18.8 g alum}`

(a) Scaling factor for 15.0 g alum

You want a theoretical yield of 15.0 g, so you must scale down the reaction.

`\text{Scale factor} = (15.0)/(18.8) = \mathbf{0.798}`

(b) Corrected volumes of NaOH and H₂SO₄

V = 0.798 × 20 mL = 16 mL