\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=0,2.0,5=0,1\left(mol\right)\)
PTHH: \(2NaAlO_2+H_2SO_4+2H_2O\rightarrow Na_2SO_4+2Al\left(OH\right)_3\)
a------->0,5a------------------------------>a
\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)
(a-0,1)---->(1,5a-0,15)
=> 0,5a + 1,5a - 0,15 = 0,1
=> a = 0,125 (mol)