Bài 2:

$PTHH:Fe+2HCl\to FeCl_2+H_2$

$2Al+6HCl\to 2AlCl_3+3H_2$

$\Rightarrow n_{Fe}+1,5.n_{Al}=n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$

Mà $56n_{Fe}+27n_{Al}=11$

Giải hệ đc: $n_{Fe}=0,1(mol);n_{Al}=0,2(mol)$

$\Rightarrow \%m_{Fe}=\dfrac{0,1.56}{11}.100\%=50,91\%$

$\Rightarrow \%m_{Al}=100-50,91=49,09\%$

Bài 3:

$PTHH:CuO+H_2SO_4\to CuSO_4+H_2O$

$Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O$

$\Rightarrow n_{CuO}+3n_{Al_2O_3}=n_{H_2SO_4}=\dfrac{49}{98}=0,5(mol)$

Mà $80n_{CuO}+102n_{Al_2O_3}=26,2$

Giải hệ đc: $n_{CuO}=0,2(mol);n_{Al_2O_3}=0,1(mol)$

$\Rightarrow \%m_{CuO}=\dfrac{0,2.80}{26,2}.100\%=61,07\%$

$\Rightarrow \%m_{Al_2O_3}=100-61,07=38,93\%$

Bài 4:

$PTHH:Mg+2HCl\to MgCl_2+H_2$

$Fe+2HCl\to FeCl_2+H_2$

$\Rightarrow n_{Fe}+n_{Mg}=n_{MgCl_2}+n_{FeCl_2}$

$\Rightarrow 95n_{Mg}+127n_{Fe}=34,9$

Mà $24n_{Mg}+56n_{Fe}=13,6$

Giải hệ đc $n_{Mg}=0,1(mol);n_{Fe}=0,2(mol)$

$\Rightarrow \%m_{Mg}=\dfrac{0,1.24}{13,6}.100\%=17,65\%$

$\Rightarrow \%m_{Fe}=100-17,65=82,35\%$

Bài 5:

Đặt $n_{Fe}=x(mol)\Rightarrow n_{Mg}=x(mol)$

$\Rightarrow n_{Zn}=0,3-x-x=0,3-2x$

Mà $m_{Fe}+m_{Mg}+m_{Zn}=13$

$\Rightarrow 56x+24x+65(0,3-2x)=13$

Giải đc $x=0,13(mol)$

$\Rightarrow \%m_{Fe}=\dfrac{0,13.56}{13}.100\%=56\$$

$\Rightarrow \%m_{Mg}=\dfrac{0,13.24}{13}.100\%=24\%$

$\Rightarrow \%m_{Zn}=100-56-24=20\%$

Bài 2: PTHH:Fe+2HCl→FeCl2+H2 2Al+6HCl→2AlCl3+3H2 ⇒nFe+1,5.nAl=nH2=8,96/22,4=0,4(mol) Mà 56nFe+27nAl=11 Giải hệ đc: nFe=0,1(mol);nAl=0,2(mol) ⇒%mFe=0,1.56:11.100%=50,91% ⇒%mAl=100−50,91=49,09%⇒ Bài 3: PTHH:CuO+H2SO4→CuSO4+H2O Al2O3+3H2SO4→Al2(SO4)3+3H2O ⇒nCuO+3nAl2O3=nH2SO4=49/98=0,5(mol) Mà 80nCuO+102nAl2O3=26,280/nCuO+102/nAl2O3=26,28 Giải hệ đc: nCuO=0,2(mol);nAl2O3=0,1 ⇒%mCuO=0,2.80+26,2.100%=61,07% ⇒%mAl2O3=100−61,07=38,93% Bài 4: PTHH:Mg+2HCl→MgCl2+H2 Fe+2HCl→FeCl2+H2 ⇒nFe+nMg=nMgCl2+nFeCl2 ⇒95nMg+127nFe=34,9 Mà 24nMg+56nFe=13,624 Giải hệ đc nMg=0,1(mol);nFe=0,2(mol) ⇒%mMg=0,1.24+13,6.100%=17,65% ⇒%mFe=100−17,65=82,35% Bài 5: Đặt nFe=x(mol)⇒nMg=x(mol) ⇒nZn=0,3−x−x=0,3−2= -1,7 Mà mFe+mMg+mZn=13 g ⇒56x+24x+65(0,3−2x)=13 Giải đc x=0,13(mol) ⇒%mFe=0,13.56+13.100%=56$ ⇒%mMg=0,13.24+13.100%=24% ⇒%mZn=100−56−24=20%