a) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PTHH: \(Al_2O_3+6HNO_3\rightarrow2Al\left(NO_3\right)_3+3H_2O\)
0,1------>0,6-------->0,2
=> \(m_{ddHNO_3}=\dfrac{0,6.63}{15\%}=252\left(g\right)\)
b) \(m_{ddspư}=252+10,2=262,2\left(g\right)\)
=> \(C\%_{Al\left(NO_3\right)_3}=\dfrac{0,2.213}{262,2}.100\%=16,25\%\)