2Al +3Cl2 -to-> 2AlCl3 (1)
vì A td vs oxi dư --> B => A: Al dư ,AlCl3
4Al +3O2 -to-> 2Al2O3 (2)
ADĐLBTKL ta có :
mO2=36,9-32,1=4,8(g)
=> nO2=0,15(mol)
theo (2) : nAl2O3=2/3nO2=0,1(mol)
=>mAl2O3(trong B)=10,2(g)
=> mAlCl3(trong B)=36,9-10,2=26,7(g)
=> mAlCl3(trong A)= 26,7(g)
theo (2) : nAl(dư)=4/3nO2=0,2(mol)
=> mAl(trong A)(dư)=5,4(g)
b)nAlCl3=26,7/133,5=0,2(mol)
theo (1) : nAl(pư)=nAlCl3=0,2(mol)
=>mAl(pư) =5,4(g)
=> m=5,4+5,4=10,8(g)