\(n_{NaOH}=0,2.2=0,4\left(mol\right)\); \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
\(n_{HCl}=\dfrac{175.14,6\%}{36,5}=0,7\left(mol\right)\)
PTHH: \(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)
0,3<-----0,3<--------------0,3<----0,45
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,1---->0,1
\(NaAlO_2+HCl+H_2O\rightarrow NaCl+Al\left(OH\right)_3\downarrow\)
0,3----->0,3---------------------->0,3
\(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)
0,1<------0,3
=> \(n_{Al\left(OH\right)_3}=0,3-0,1=0,2\left(mol\right)\Rightarrow m=0,2.78=15,6\left(g\right)\)