Question

Answer 1

Bài 6:

Đặt nFe(p.ứ)=a(mol)

PTHH 3 Fe + 2 O2 -to-> Fe3O4

a________2/3a____1/3a(mol)

Ta có: m(rắn)= mFe(dư)+ mFe3O4

<=> 28,8= (22,4-56a)+ 1/3a.232

<=>64/3 a= 6,4

<=>a=0,3(mol)

=> mFe(p.ứ)=0,3.56=16,8(g)

=>H(p.ứ)= (16,8/22,4).100=75%

Answer 2

Bài 7:

nAl=5,4/27=0,2(mol)

nCr2O3= 15,2/152=0,1(mol)

PTHH: 2 Al + Cr2O3 -to-> Al2O3 + 2 Cr

Ta có: 0,2/2 = 0,1/1

-> P.Ứ hết 

=> nCr(LT)=nAl=0,2(mol)

Vì: H=75% => nCr(TT)=0,2.75%= 0,15(mol)

=>mCr= 52. 0,15=7,8(g)