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

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

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

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: \(MgSO_4;Al_2\left(SO_4\right)_3\)

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

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

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

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

\(a.n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+24b=5,1\\1,5a+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,1}.100\approx52,941\%\\\%m_{Mg}\approx47,059\%\end{matrix}\right.\)

\(b.m_{ddH_2SO_4}=\dfrac{0,25.98.100}{9,8}=250\left(g\right)\\ m_{ddsau}=m_{Al,Mg}+m_{ddH_2SO_4}-m_{H_2}=5,1+250-0,25.2=254,6\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342}{254,6}.100\approx6,716\%\\ C\%_{ddMgSO_4}=\dfrac{0,1.120}{254,6}.100\approx4,713\%\)

bC

a)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:       x                                                     1,5x

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:      y                                                 y

Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)

b) 

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:      0,1      0,15                  0,05                            

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:     0,1       0,1                 0,1

\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)

m_{dd sau pứ} = 5,1+250-0,15.2 = 254,8(g)

\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)

\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)

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

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

b) Gọi x,y là số mol Al, Fe

\(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

Ta có hệ : \(\left\{{}\begin{matrix}27x+56y=0,83\\\dfrac{3}{2}x+y=0,02\end{matrix}\right.\)

=> \(x=\dfrac{29}{5700};y=\dfrac{47}{3800}\)

\(\%m_{Al}=\dfrac{\dfrac{27}{5700}.27}{0,83}.100=16,55\%\); \(\%m_{Fe}=100-16,55=83,45\%\)

c)Bảo toàn nguyên tố H: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)

=> \(C\%_{H_2SO_4}=\dfrac{0,02.98}{200}.100=0,98\%\)

a) Gọi số mol Mg, CuO là a, b (mol)

=> 24a + 80b = 14 (1)

\(n_{HCl}=\dfrac{255,5.10\%}{36,5}=0,7\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a--->2a--------->a------>a

           CuO + 2HCl --> CuCl₂ + H₂O

               b------>2b----->b

=> 2a + 2b = 0,7 (2)

(1)(2) => a = 0,25 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{14}.100\%=42,857\%\\\%m_{CuO}=\dfrac{0,1.80}{14}.100\%=57,143\%\end{matrix}\right.\)

b)

m_{dd sau pư} = 14 + 255,5 - 0,25.2 = 269 (g)

\(C\%_{MgCl_2}=\dfrac{0,25.95}{269}.100\%=8,829\%\)

\(C\%_{CuCl_2}=\dfrac{0,1.135}{269}.100\%=5,019\%\)

ko biết viết thì đừng có nhắn

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

b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

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

_____0,02<---0,03<---------------------0,03

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)

c) m_H2SO4 = 0,03.98 = 2,94 (g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)

PTHH: 

Zn + H₂SO₄ ---> ZnSO₄ + H₂ (1)

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂ (2)

Ta có: \(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

Gọi x, y lần lượt là số mol của Zn và Al

a. Theo PT₍₁₎: \(n_{H_2}=n_{Zn}=x\left(mol\right)\)

Theo PT₍₂₎: \(n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}y\left(mol\right)\)

=> \(x+\dfrac{3}{2}y=0,8\) (*)

Theo đề, ta có: 65x + 27y = 3,79 (**)

Từ (*) và (**), ta có HPT:

\(\left\{{}\begin{matrix}x+\dfrac{3}{2}y=0,8\\65x+27y=3,79\end{matrix}\right.\)

(Ra số âm, bn xem lại đề nhé.)