HOC24
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nAl=8,1/27=0,3
nHCl=21,9/36,5=0,6
2 Al+6HCl→2AlCl3+3H2
trước pư: 0,3 0,6 (mol)
Pư : 0,2 ← ,6 → 0,2 (mol)
sau pư : 0,1 0 0,2 (mol)
→Al dư HCl pư hết
mAl (dư)=0,1*27=2,7(g)
mAlCl3=0,2*133,5=26,7
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h%=mtt/mlt*100
ct này bn tính chất tham gia pư
vd Fe+O2→Fe3O4
Fe,O2 la chấttham gia pư
Fe3O4 là sp
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