The aluminum in a 1.200-g sample of impure ammonium aluminum sulfate was precipitated with aqueous ammonia as Al2O3*H20. The precipitate was filtered and ignited at 1000 degrees C to give anhydrous Al2O3, which had a mass of 0.1798g.
a) What is the percentage of Al in the original sample?
b) What is the percentage of NH4Al(SO4)2 in the original sample?
2NH4Al(SO4)2 = 2NH3 +Al2O3.H2O + 4SO2 + 2O2
Atomic mass of
Al = 27
S = 32
H = 1
O = 16
N = 14
Molecular mass of
NH4Al(SO4)2 = 14 +4 +27 ...
The solution lays out all calculations necessary to find the percentage of aluminum and ammonium aluminum sulfate in a sample before being filtered and ignited when given the conditions of the reaction and the products.