a) PTHH:3NaOH+AlCl3-->Al(OH)3+3NaCl
b) Ta có C%=mct/mdd*100
=>mctAlCl3=C%*mdd/100=20*6,675/100=1,335g
nAlCl3=m/M=1,335/133,5=0,01mol
mà nNaOH=3nAlCl3=3*0,01=0,03mol
cM=n/V=>VNaOH=n/cM=0,03/1=0,03l
c) nAl(OH)3=nAlCl3=0,01mol
=>mAl(OH)3=n*M=0,01*78=0,78g