PTHH: \(3NaOH+Al\left(NO_3\right)_3\rightarrow3NaNO_3+Al\left(OH\right)_3\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\cdot1=0,2\left(mol\right)\\n_{Al\left(NO_3\right)_3}=0,5\cdot0,2=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,1}{1}\) \(\Rightarrow\) Al(NO3)3 còn dư, tính theo NaOH
\(\Rightarrow n_{Al\left(OH\right)_3}=\dfrac{1}{15}\left(mol\right)\) \(\Rightarrow m_{Al\left(OH\right)_3}=\dfrac{1}{15}\cdot78=5,2\left(g\right)\)