OWLv2 lOnline teaching and learning resource from Cengage Learning-Google Chrome
ID: 483157 • Letter: O
Question
OWLv2 lOnline teaching and learning resource from Cengage Learning-Google Chrome O east cen take CovalentActivity do?locato assignment take&takeAssignmentsessionLocator; assignment take Chapter 12: EOC The aluminum in a 1.328-g sample of impure ammonium aluminum sulfate was precipitated with aqueous ammonia as the hydrous Al2Os H2O. The precipitate was filtered and Question 1 1 pt ignited at 1000°C to give anhydrous Al203, which weighed 0.1880 g. Express the result of this analysis in tems of Question 2 1pt a. NH Al (SO Question 3 1 pt b. Al20 Question 4 1 Pt Question 5 1 pt Question 6 1pt Submit Answer Try Another Version 3 item attempts remaining Question 7 1 pt Progress 07 items Due Feb 14 at Previous 05:00 PM Finish Assig nment nstructor Save and Exit Cengage Learning l Cengage Technical Support 5:35 PM a Search XExplanation / Answer
MW = molecular weight
(a) % NH4Al(SO4)2
Mass of NH4Al(SO4)2 = mass Al2O3 x (2 * MW of NH4Al(SO4)2 / MW of Al2O3)
= 0.188 g x (2*237.15/101.96)
= 0.8745 g g
% NH4Al(SO4)2 = (mass NH4Al(SO4)2 / total mass of the sample ) x 100
= (0.8745 g/1.328 g) x 100
= 65.8 %
Therefore,
% NH4Al(SO4)2 = 65.8 %
(b) % Al2O3
% Al2O3 = (mass of Al2O3 / total mass of the sample ) x 100
= (0.188 g/1.328 g) x 100%
= 14.15 %
Therefore,
% Al2O3 = 14.15 %
(c) % Al
Mass of Al = mass Al2O3 x (2 x Al / MW of Al2O3)
= 0.188 g x (2x 26.98 g/101.96 g)
= 0.1 g
Then,
% Al = ( mass of Al/ total mass of the sample) x 100
= ( 0.1 g / 1.328 g ) x 100
= 7.53 %
Therefore,
% Al = 7.53 %
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