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.

(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

(ii) Moles of alum

(iii) Theoretical yield of 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.

(b) Corrected volumes of NaOH and H₂SO₄

V = 0.798 × 20 mL = 16 mL

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2