Question

Answer 1

Câu 6:

nAl=3,24/27=0,12(mol); nO2= 4,48/22,4=0,2(mol)

PTHH: 4 Al + 3 O2 -to-> 2 Al2O3

Ta có: 0,12/4 < 0,2/3

=> O2 dư, Al hết, tính theo nAl

=> nAl2O3(LT)= nAl/2= 0,12/2=0,06(mol)

nAl2O3(TT)=4,59/102=0,045(mol)

=> H= (0,045/0,06).100= 75%

Answer 2

Câu 7:

nMg=6/24=0,25(mol); nS= 8,8/32=0,275(mol)

PTHH: Mg + S -to-> MgS

Ta có: 0,25/1 < 0,275/1

=> Mg hết, S dư, tính theo nMg

=> nMgS(LT)=nMg= 0,25(mol)

nMgS(TT)= 10,08/56= 0,18(mol)

=>H= (0,18/0,25).100=72%

Answer 3

Câu 8:

nZn= 12,35/65=0,19(mol)

nO2=1,792/22,4=0,08(mol)

PTHH: 2 Zn + O2 -to->  2 ZnO

Ta có: 0,19/2 > 0,08/1

-> Zn dư, O2 hết, tính theo nO2

Ta có: nZnO(LT)= nO2.2=0,08.2=0,16(mol)

nZnO(TT)= 9,72/81=0,12%

=> H=(0,12/0,16).100=75%

Answer 4

Câu 9:

nAl= 3,51/27=0,13(mol)

nO2=2,24/22,4=0,1(mol)

PTHH: 4 Al + 3 O2 -to-> 2 AlO3

nAl2O3(TT)=5,1/102=0,05(mol)

Ta có: 0,13/4 < 0,1/3

=> Al hết, O2 dư, tính theo nAl

=> nAl2O3(LT)=0,13/2=0,065(mol)

=>H=(0,05/0,065).100=76,923%