nAl=275,4=0,2(mol)
PTHH:
2Al+6HCl→2AlCl3+3H2
a,
Theo PTHH, nHCl=3nAl=0,6(mol)
→mHCl=0,6.36,5=21,9g
b,
Theo PTHH, nH2=23nAl=0,3(mol)
→VH2=0,3.22,4=6,72l
c,
+ Cách 1:
Theo PTHH, nAlCl3=nAl=0,2(mol)
→mAlCl3=0,2.133,5=26,7g
+ Cách 2:
mH2=0,3.2=0,6g
BTKL:
mAlCl3=mAl+mHCl−mH2=5,4+21,9−0,,6=26,7g