\(n_{AlCl_3}=\dfrac{350.13,35\%}{133,5}=0,35\left(mol\right)\)
\(n_{Na}=\dfrac{27,6}{23}=1,2\left(mol\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
1,2----------------->1,2----->0,6
\(3NaOH+AlCl_3\rightarrow Al\left(OH\right)_3+3NaCl\)
1,05<----0,35------>0,35------>1,05
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
0,15<-----0,15------->0,15
\(\left\{{}\begin{matrix}m_{NaCl}=1,05.58,5=61,425\left(g\right)\\m_{NaAlO_2}=0,15.82=12,3\left(g\right)\\m_{Al\left(OH\right)_3}=\left(0,35-0,15\right).78=15,6\left(g\right)\end{matrix}\right.\)
mdd sau pư = 27,6 + 350 - 0,6.2 - 15,6 = 360,8 (g)
\(\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{61,425}{360,8}.100\%=17,025\%\\C\%_{NaAlO_2}=\dfrac{12,3}{360,8}.100\%=3,409\%\end{matrix}\right.\)
