a) PTHH:3NaOH+AlCl3\(\rightarrow\)Al(OH)3+3NaCl
b) Ta có
mAlCl3==\(\frac{\text{20.6,675}}{100}\)=1,335g
nAlCl3=\(\frac{1,335}{133,5}\)=0,01mol
mà nNaOH=3nAlCl3=3.0,01=0,03mol
\(\rightarrow\)VNaOH=\(\frac{0,03}{1}\)=0,03l
c) nAl(OH)3=nAlCl3=0,01mol
\(\rightarrow\)mAl(OH)3=0,01.78=0,78g