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

               2a______3a__________a_______3a    (mol)

            \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

                b_______b________b______b        (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27\cdot2a+24b=7,8\\3a+b=\dfrac{200\cdot19,6\%}{98}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1\cdot24=2,4\left(g\right)\\m_{Al}=5,4\left(g\right)\\n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)=n_{MgSO_4}\\n_{H_2}=0,4\left(mol\right)\Rightarrow m_{H_2}=0,4\cdot2=0,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=207\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1\cdot342}{207}\cdot100\%\approx16,52\%\\C\%_{MgSO_4}=\dfrac{0,1\cdot120}{207}\cdot100\%\approx5,8\%\end{matrix}\right.\)

Bài 3 : 

a) $Mg + H_2SO_4 \to MgSO_4 + H_2$
$n_{Mg} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$\%m_{Mg} = \dfrac{0,15.24}{13,2}.100\% = 27,27\%$
$\%m_{Cu} = 100\% -27,27\% = 72,73\%$

b) $n_{Cu} = \dfrac{13,2 - 0,15.24}{64}= 0,15(mol)$

$\Rightarrow m_{muối} = 0,15.120 + 0,15.160= 42(gam)$

Bài 4 : 

Gọi $n_{Fe} = a(mol) ; n_{Mg} = b(mol)$
$56a + 24b = 18,4(1)$

$Fe + 2HCl \to FeCl_2 + H_2$
$Mg + 2HCl \to MgCl_2 + H_2$

Theo PTHH : $n_{H_2} = a + b = \dfrac{11,2}{22,4} = 0,5(2)$

Từ (1)(2) suy ra a = 0,2 ; b = 0,3

$\%m_{Fe} = \dfrac{0,2.56}{18,4}.100\%  = 60,87\%$

$\%m_{Mg} = 100\% -60,87\% = 39,13\%$

b) $n_{HCl} = 2n_{H_2} = 1(mol)$
$V_{dd\ HCl} = \dfrac{1}{0,8}=  1,25(lít)$

3a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(n_{H_2}=n_{Mg}=0,15\left(mol\right)\)

=> \(\%m_{Mg}=\dfrac{0,15.24}{13,2}.100=27,28\%\)

\(\%m_{Cu}=100-27,28=72,73\%\)

b) \(n_{Cu}=\dfrac{13,2-3,6}{64}=0,15\left(mol\right)\)

Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

0,15------------------->0,15

Mg + 2H2SO4 → MgSO4 + SO2 + 2H2O

0,15------------------->0,15

\(m_{muối}=0,15.160+0,15.120=42\left(g\right)\)

a, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

m_{hh Zn và Fe} = 21,6-3 = 18,6 (g)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:      x                                           x

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:      y                                           y 

Ta có: \(\left\{{}\begin{matrix}65x+56y=18,6\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%m_{Zn}=\dfrac{0,2.65.100\%}{21,6}=60,19\%\)

\(\%m_{Fe}=\dfrac{0,1.56.100\%}{21,6}=25,93\%\)

\(\%m_{Cu}=100-60,19-25,93=13,88\%\)

b,

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,2     0,2                            0,2

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,1      0,1                            0,1 

\(m_{H_2SO_4}=\left(0,1+0,2\right).98=29,4\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4.100\%}{25\%}=117,6\left(g\right)\)

c,m_{dd sau pư} = 21,6+117,6- (0,1+0,2).2 = 138,6 (g)

\(C\%_{ddZnSO_4}=\dfrac{0,2.161.100\%}{138,6}=23,23\%\)

\(C\%_{ddFeSO_4}=\dfrac{0,1.152.100\%}{138,6}=10,97\%\)

Bài 1:

a+b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

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

\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\) \(\Rightarrow m_{Cu}=6,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{17,6}\cdot100\%\approx63,64\%\\\%m_{Cu}=36,36\%\end{matrix}\right.\)

c) Ta có: \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

Bảo toàn nguyên tố: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)=n_{CuSO_4}\)

\(\Rightarrow m_{muối}=0,1\cdot400+0,1\cdot160=56\left(g\right)\)

Bài 2:

Quy đổi hh gồm Fe (a mol) và O (b mol)

\(\Rightarrow56a+16b=27,6\)  (1)

Ta có: \(n_{SO_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

Bảo toàn electron: \(3n_{Fe}=2n_O+2n_{SO_2}\) \(\Rightarrow3a-2b=0,45\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,39\\b=0,36\end{matrix}\right.\)

Bảo toàn nguyên tố: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,195\left(mol\right)\) \(\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,195\cdot400=78\left(g\right)\)

\(m_{H_2SO_4}=\frac{200.19,6}{100}=39,2g\)

\(\rightarrow n_{H_2SO_4}=\frac{39,2}{98}=0,4mol\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

x                  x                x                    x

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

y               1,5y                    0,5y                 1,5y

Đặt \(\hept{\begin{cases}x\left(mol\right)=Mg\\y\left(mol\right)=Al\end{cases}}\)

a. Có hệ phương trình là: \(\hept{\begin{cases}24x+27y=7,8\\x+1,5y=0,4\end{cases}}\)

\(\rightarrow\hept{\begin{cases}x=0,1\\y=0,2\end{cases}}\)

\(m_{Mg}=0,1.24=2,4g\)

\(m_{Al}=7,8-2,4=5,4g\)

b. \(n_{H_2}=0,1+1,5.0,2=0,4mol\)

\(\rightarrow m_{H_2}=0,4.2=0,8g\)

\(m_{ddsaupu}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}\)

\(\rightarrow m_{ddsaupu}=7,8+200-0,8=207g\)

Những dung dịch thu được sau phản ứng: \(MgSO_4;Al_2\left(SO_4\right)_3\)

\(m_{MgSO_4}=0,1.\left(24+32+16.4\right)=12g\)

\(\rightarrow C\%_{MgSO_4}=\frac{12.100}{207}=5,8\%\)

\(m_{Al_2\left(SO_4\right)_3}=0,5.0,2.\left(27.2+32.3+16.12\right)=34,2g\)

\(\rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{34,2.100}{207}=16,52\%\)

nH2=2,24/22,4=0,1(mol)

PTHH:

Fe + H2SO4 - FeSO4 + H2

1    -    1        -     1        -   1          (mol)

0,1  -    0,1     -    0,1      -    0,1      (mol)

Theo PTHH, ta có:

nFe=0,1 (mol)

mFe=0,1*56=5,6(g)

%mFe=5,6/13 *100=43%

mCu=mhh-mFe=13-5,6=7.4(g)

%mCu=7.4/13 *100=56,9%

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2,24 lít khí chứ k phải 22,4