Question

Calculate the % Al in the sample.

Preparation of a stock solution of aluminum. Accurately weigh out approximately 1 g of the aluminum foil. Tear it up into small pieces, place the pieces into a covered 250 mL beaker and add 60 mL of 6M HCl. This should be carried out in the fume hood because the reaction is exothermic and leads to violent frothing and emission of vapors. Cool, quantitatively transfer to a 250 mL flask, rinsing with deionized water, fill up to the mark and mix thoroughly. Label this stock solution accordingly.

Procedure for Method 2. Pipette 10 mL of the stock Al solution into a 400 mL beaker and carefully add dilute ammonia solution until precipitation just takes place. Add dilute HCl acid until the precipitate just redissolves. Dilute to 50 mL with water, warm to 70 – 80 °C, and add slowly with stirring a slight excess of the oxine reagent. Add 10 mL of 70% ammonium acetate solution to ensure complete precipitation of the complex. The supernatant liquid should be yellow if excess is present. Allow this to stand for 30 minutes on a warm hot plate (not boiling) without stirring. Allow mixture to cool (possibly overnight). Vacuum filter the precipitate through a previously dried and weighed No. 4 sintered glass funnel. Wash the precipitate with water and dry in air and then in an oven at 120 – 140 °C to constant weight.

Calculate the % Al in the sample.

Mass of the Al foil used for the stock solution is 1.0241 g

Mass of precipitate is 0.6414 g

The following equations may be helpful?

Al(OH)3 ⇌ Al(OH)4^-

NH4Cl → NH4 + + Cl−

NH4 + + H2O ⇌ NH4OH + H+

2 Al(OH)3 → Al2O3.xH2O → γ Al2O3 → αAl2O3

Answer 1

The the objective is to calculate the sample to the following data. "Al in Aus Given that, The related equations for the formation of Al2O3 from Al is given below. . ALCS) - > Alyt (aq) +36 ------- C1) ALS (aq) of 3 0H (9) -> Al fots (3) ---- (2) 2ALCOH), (5) — Al2O3 - 3H2015) -----(3) Al2O3. 3H₂O (3) Al2O3 (3) + 3H₂O --o0 (4) As per the Stoichiometric equations above, i mol AL = 1 mol AL COH)2. 2 mols ALCOH)3 = 1 mol Al2O3, The atomie masses are ; AL = 26.981 g/mol O $ 15999 g/mol.

The gram molar mass of Al2O3= = (2x26.981 +3815.999) 3/mol € (53,962 & 47.997) g/mol 101.959 g/mol. Mass of the residue = 0.6414 g. Mo (5) Al2Os corresponding to 0641495 :-(0.64149)/(101.959 g/mol) -0.006291 mol Mol (3) Al corresponding to 0.006291 mol Alacasa -(0.006291 mol Alzcog) *(z mols AL) /como Alzcom] < 0.012582 mol. Mass of Al in the foil corresponding to 0.012582 mol =(0.012582 molyX(26.981 glmoy R = 0.33947 9 ñ 0.33959 - Mass percent Al in the foil = =(0.33959) (1.02419) X100 =133.15))