PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

              a_______a_______a_____a    (mol)

            \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

                2b______3b__________b_____3b    (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27\cdot2b=11\\a+3b=0,2\cdot2=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{11}\cdot100\%\approx50,91\%\\\%m_{Al}=49,09\%\end{matrix}\right.\)

Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{FeSO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\) 

a) nH2SO4=0,4(mol)

Đặt: nFe=x(mol); nAl=y(mol) (x,y>0)

PTHH: Fe + H2SO4 -> FeSO4 + H2

x________x______x______x(mol)

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

y____1,5y_______0,5y_______1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=> mFe=0,1.56=5,6(g)

=>%mFe=(5,6/11).100=50,909%

=>%mAl= 49,091%

b) V(H2,đktc)=0,4.22,4=8,96(l)

c) nAl2(SO4)3= 0,5y=0,5.0,2=0,1(mol)

nFeSO4=x=0,1(mol)

Vddsau=VddH2SO4=0,2(l)

=>CMddAl2(SO4)3= 0,1/0,2=0,5(M)

CMddFeSO4=0,1/0,2=0,5(M)

\(n_{HCl}=\frac{182,5.10}{100.36,6}=0,5\left(mol\right)\)

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Theo PT: \(n_{Fe}=n_{H_2}=\frac{1}{2}n_{HCl}=0,25\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)

\(\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

b) Áp dụng ĐLBTKL nên ta có:

\(m_{ddFeCl_2}=m_{Fe}+m_{ddHCl}-m_{H_2}\)

\(\Leftrightarrow m_{ddFeCl_2}=14+182,5-0,25.2=196\left(g\right)\)

c)Theo PT: \(n_{FeCl_2}=\frac{1}{2}n_{HCl}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{FeCl_2}=\frac{0,25.127}{196}.100=16,2\%\)

\(n_{HCl}=\frac{10.182,5}{100.36,5}=0,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ m_{Fe}=56.0,25=14\left(g\right)\\ V_{H_2}=22,4.0,25=5,6\left(l\right)\\ m_{ddA}=14+182,5-0,25.2=196\left(g\right)\\ m_A=0,25.127=31,75\left(g\right)\\ C\%_{ddA}=\frac{31,75}{196}.100\%=16,2\left(\%\right)\)

a) \(n_{HCl}=\frac{182,5.10}{100.36,5}=0,5mol\)

PTHH : \(Fe+2HCl\rightarrow FeCl_2+H_2\)

( mol ) 0,25 0,5 0,25 0,25

=> mFe = 56 . 0,25 = 14 g

V_H2 = 0,25 . 22,4 = 5,6 l

b) Áp dụng ĐLBTKL ta có :

mdd FeCl2 ( A) = 14 + 182,5 - 0,25 . 2 =196 g

c) mFeCl2 ( A) = 0,25 . 127 = 31,75g

=>C% dd FeCl2(A) = \(\frac{31,75}{196}.100\%=16,2\%\)

n_Mg = 0,1(mol)

PTHH: Mg + 2HCl --> MgCl₂ +H₂

n_Mg = n_MgCl2= n_H2 = 0,1(mol)

=> m_muối = 9,5(g) 

_VH2 = 2,24(l)

b) CM_HCl = 0,2/0,1=2(M)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2\cdot56=11,2\left(g\right)\\m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=156,8\left(g\right)\) \(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{156,8}\cdot100\%\approx16,2\%\)

Theo gt ta có: $n_{Al}=0,1(mol)$

a, $2Al+6HCl\rightarrow 2AlCl_3+3H_2$

b, $\Rightarrow n_{H_2}=0,15(mol)\Rightarrow V_{H_2}=3,36(l)$

c, Ta có: $n_{HCl}=0,3(mol)\Rightarrow m_{HCl}=10,95(g)\Rightarrow \%m_{ddHCl}=219(g)$

d, Bảo toàn khối lượng ta có: $m_{dd}=221,4(g)$

$\Rightarrow \%C_{AlCl_3}=6,02\%$

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

Fe + H2SO4 → FeSO4 + H2 

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Gọi x,y lần lượt là số mol Fe, Al

\(\left\{{}\begin{matrix}56x+27y=11\\x+\dfrac{3}{2}y=0,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=>\(\%m_{Fe}=\dfrac{0,1.56}{11}.100=50,91\%\)

=> %m _Al = 100 - 50,91 =49,09 %

b)Theo PT:  \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)

=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c) \(CM_{FeSO_4}=\dfrac{0,1}{0,2}=0,5M\)

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,2}{2}}{0,2}=0,5M\)