mCu=mCuO/MCuO*MCu=1.1g=>nAl*MAl+nMg*MMg=5-1.1=3.9
neu hoc ppe thi theo ppe them 1 pt giai he pt 2an ; neu chua hoc ghi pt hoa hoc can bang ra dc 1 pt nua
nAl=0.1=>m=2.7;nMg=0.05g=>m=1.2g
tu pt co the biet dc tong so mol cua HCl;theo ppe thi dung dlbt ngtu ngto H
=>nHCl=0.4mol=>mHCl=nM=14.6g=>mdd=mHCl*100/C%=192.105....g
BTKL: mdds =mddHCl+mAl+mMg=196.005g
tu pt hoac bao toan ngtu ngto nAlCl3=0.1;nMgCl2=0.05=>C%AlCl3=(M*n/mdds)*100=3.18869%
C%MgCl2=(M*n/mdds)*100=2.423%
bua sau nen ghi lop may