Question

Giúp mình 2 câu này với

Answer 1

Bài 15:

a) Zn + 2 HCl -> ZnCl2 + H2

2 Al+ 6 HCl -> 2 AlCl3 + 3 H2

Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O

Al(OH)3 +3 HCl -> AlCl3 +3 H2O

Fe3O4 + 8 HCl -> FeCl2 +2 FeCl3 +4 H2O

Fe +2 HCl -> FeCl2 + H2

Fe(OH)3 +3 HCl -> FeCl3 + 3 H2O

CaCO3 +2 HCl -> CaCl2 + CO2 + H2O

AgNO3 + HCl -> AgCl + HNO3

Answer 2

Bài 15b)

Ba(OH)2 + H2SO4 -> BaSO4 +2 H2O

BaCl2 + H2SO4 -> BaSO4 + 2 HCl

Cu(OH)2 + H2SO4 -> CuSO4 + 2 H2O

CuO + H2SO4 -> CuSO4 + H2O

Fe2O3 + 3 H2SO4 -> Fe2(SO4)3 +3 H2O

Fe + H2SO4 -> FeSO4 + H2

Fe(OH)2 + H2SO4 -> FeSO4 + 2 H2O

2 Fe(OH)3 + 3 H2SO4 -> Fe2(SO4)3 + 6 H2O

2Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2

Al2O3 + 3 H2SO4 -> Al2(SO4)3 +3 H2O

Answer 3

Câu 14:

a) Đặt nZn=a(mol); nFe=b(mol) (a,b >0)

nH2SO4 = 0,5(mol); nH2=0,3(mol)

Vì: 0,5/1 > 0,3/1 => H2SO4 dư, hỗn hợp kim loại dùng hết

PTHH: Zn + H2SO4 -> ZnSO4 + H2

a_________a______a______a(mol)

Fe+ H2SO4 -> FeSO4 + H2

b____b___b______b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}56a+65b=17,7\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mFe=0,2.56=11,2(g)

=>%mFe= (11,2/17,7).100=63,277%

=>%mZn= 36,723%

b) nFeSO4= nFe=0,2(mol); nZnSO4=nZn=0,1(mol)

nH2SO4(dư)=0,5-0,3=0,2(mol)

Vddsau= VddH2SO4(bđ)=0,5(l)

=>CMddH2SO4(dư)=0,2/0,5=0,4(M)

CMddZnSO4= 0,1/0,5=0,2(M)

CMddFeSO4=0,2/0,5=0,4(M)