a) Đặt: nFe= x(mol); nZn=y (mol) (x,y: nguyên, dương)
PTHH: Fe + 2 HCl -> FeCl2 + H2
x________________x______x(mol)
Zn+ 2 HCl -> ZnCl2 + H2
y____________y___y(mol)
b)
Ta được hpt: \(\left\{{}\begin{matrix}56x+65y=18,6\\22,4x+22,4y=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=> mFe= 0,1.56=5,6(g); mZn=0,2.65= 13(g)
$(1)CaCO_3 \xrightarrow{t^o} CaO + CO_2$
$(2) CaO + H_2O \to Ca(OH)_2$
$(3) Ca(OH)_2 + CO_2 \to CaCO_3 + H_2O$
$(4) CaO + 2HCl \to CaCl_2 + H_2O$
$(5) Ca(OH)_2 + 2HNO_3 \to Ca(NO_3)_2 + 2H_2O$
\(4FeS_2+11O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3+8SO_2\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}3Fe+3H_2O\)
\(2Fe+6H_2SO_{4\left(đ\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(Fe_2\left(SO_4\right)_3+3BaCl_2\rightarrow2FeCl_3+3BaSO_4\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}3Fe+3H_2O\)
\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)
$4FeS_2 +11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
$2Fe + 6H_2SO_{4_{đ}} \xrightarrow{t^o} Fe_2(SO_4)_3 + 3SO_2 + 6H_2O$
$Fe_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2FeCl_3$
$FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$
$2Fe + 3Cl_2 \to 2FeCl_3$
Chuỗi 2:
(1) 4 FeS2 +11 O2 -to,xt-> 2 Fe2O3 + 8 SO2
(2) Fe2O3 + H2 -to-> 2 Fe + 3H2O
(3) 2 Fe + 6 H2SO4(đ) -to-> Fe2(SO4)3 + 3 SO2 +6 H2O
(4) Fe2(SO4)3 + 3 BaCl2 -> 2 FeCl3 + 3 BaSO4
(5) FeCl3 + 3 NaOH -> Fe(OH)3 + 3 NaCl
(6) 2 Fe(OH)3 -to-> Fe2O3 + 3 H2O
(7) Fe2O3 + 3 CO -to-> 2 Fe + 3 CO2
(8) 2 Fe + 3 Cl2 -to-> 2 FeCl3
