lab6 - up The precipitate soon dissolved again as the last...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Section: I2 Date of experiment performed: 01/26/2009 Date Report prepared: 01/31/2009 Name: Fletcher Dostie Partner's Name: Andrew Pak Experiment 6 : Synthesis of Alum from Al soda Can Data/Calculations: (60 pts total) (20 pts) MW moles Limiting? 100% Yield (g) % yield Potassium hydroxide, 1.4 M KOH 56.10 50.00 0.070 No Aluminum Al 26.98 1.073 0.040 Yes Sulfuric Acid, 9M 98.08 20.00 0.180 No Alum 474.39 11.369 18.87 60.26 ml or g used/made H 2 SO 4 K[Al(SO 4 ) 2 ] . 12 H 2 O OBSERVATIONS for Synthesis , Separation & Crystallization (20 pts) When the aluminum was added to 50 ml of 1.4 M KOH and the mixture was placed on a hot plate, the mixture turned black and bubbled, and the aluminum periodically rose and fell. This could be due to hydrogen bubbles forming on the aluminum fragments, which caused them to float in the solution, lifting them to the top. The bubbles would then pop and the aluminum would sink again. When filtered, the solution became clear. Adding 20 ml of 9.0 M H 2 SO 4 caused a white precipitate to form, and the solution to heat
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: up. The precipitate soon dissolved again as the last of the H 2 SO 4 was added. The solution was then chilled to allow alum crystals to form. However, addition of 95% ethanol was needed to lower the alum solubility and allow it to crystallize from the solution. After filtering and drying the alum, it appeared as a fluffy white powder. 2 Sample Equations: (20 pts-5 pts each) Limiting Reagent: Al is the limiting reagent. 100% yield % yield NOTE: Use this area to show sample equations and calculations. Use Equation Editor. Moles of H 2 SO 4 used 9.0 moles 1L ⋅ 0.020 L= 0.18 moles 1.073 gAl ⋅ 1 moleAl 26.98 g = 0.040 moles 1.4 moles 1L ⋅ 0.050 L= 0.070 moles 0.040 moles / 2 = 0.020 moles 0.070 moles / 2 = 0.035 moles 0.180 moles / 4 = 0.045 moles 0.040 moles ⋅ 474.39 g 1 mole = 18.87 g 11.369 g 18.87 g ⋅ 100 = 60.26...
View Full Document

This note was uploaded on 04/16/2009 for the course CHEM 1312 taught by Professor Bottomley during the Spring '07 term at Georgia Tech.

Page1 / 2

lab6 - up The precipitate soon dissolved again as the last...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online