a) 2Al+3H2SO4---->Al2(SO4)3 +3H2(1)
Al2O3 +3H2SO4--->Al2(SO4)3 +3H2O(2)
Ta có
n\(_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Theo pthh
n\(_{Al}=\frac{2}{3}n_{H2}=0,2\left(mol\right)\)
m\(_{Al}=0,2.27=5,4\left(g\right)\)
%m\(_{Al}=\)\(\frac{5,4}{15,6}.100\%=34,61\left(\%\right)\)
%m\(_{Al2O3}=100\%-34,61\%=65,39\%\%\)
b)Theo pthh1
n\(_{Al2\left(SO4\right)3}=\frac{1}{3}n_{H2}=0,1\left(mol\right)\)
m\(_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
Ta có
m\(_{Al2O3}=15,6-5,4=10,2\left(g\right)\)
=> n\(_{Al2O3}=\frac{10,2}{102}=0,1\left(mol\right)\)
Theo pthh2
n\(_{Al2\left(SO4\right)3}=n_{Al2O3}=0,1\left(mol\right)\)
m\(_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
m\(_{Al2\left(SO4\right)3}=34,2+34,2=68,4\left(g\right)\)
C%=\(\frac{68,4}{500}.100\%=13,68\%\)
