$mMuối=mBr_{2}+mCl_{2}+mKL$
⇔$9,4=mBr_{2}+mCl_{2}+0,54+0,24+0,8$
⇒$mBr_{2}+mCl_{2} =7,82$
⇒$160nBr_{2}+71nCl_{2}=7,82(1)$
$nAl=\frac{0,54}{27}=0,02$
$nMg=\frac{0,24}{24}=0,01$
$nCa=\frac{0,8}{40}=0,02$
Bảo toàn e ta có:
$2nBr_{2}+2nCl_{2}=3nAl+2nMg+2nCa$
⇒$2nBr_{2}+2nCl_{2}=3.0,02+2.0,01+2.0,02=0,12(2)$
(1)(2)⇒$\left \{ {{nBr_{2}=0,04} \atop {nCl_{2}=0,02}} \right.$
$mBr_{2}=0,04.160=6,4g$
$mCl_{2}=0,02.71=1,42g$
⇒$A$