1)Al+4HNO3-->Al(NO3)3+NO+2H2O 
x-----------4x----------------------x 
Al2O3+6HNO3-->2Al(NO3)3+3H2O 
y----------6y 
x=0,6=>m Al=16,2=>% Al=76,05% 
=>m Al2O3=5,1=>% Al2O3=23,9% =>y=0,05 
=>n HNO3=0,6x4+0,05x6=2,7 mol 
=>V=2,7/2=1,35M 
=>n Al(NO3)3=0,7 
n Al(OH)3=31,2/78=0,4 
Al(NO3)3+3NaOH-->Al(OH)3+3NaNO3 
0,7----------------------------0,7 
=.>Al(OH)3+ NaOH-->NaAlO2+3H2O 
tạo thành 0,4 mol kết tủa nên Al(OH)3 tan bớt 0,7-0,4=0,3 mol 
=>n NaOH=0,3 
=> V NaOH=0,4/2=0,2l =200ml