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

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

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)

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

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,1mol\\ m_{Zn}=0,1.65=6,5g\\ m_{Cu}=9,7-6,5=3,2g\\ b)C_{\%ZnCl_2}=\dfrac{0,1.136}{6,5+120-0,1.2}\cdot100=10,77\%\)

\(a)n_{Mg} = a ; n_{Al} = b \Rightarrow 24a +27b = 5,1(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{5,6}{22,4} = 0,25(2)\\ (1)(2) \Rightarrow a = 0,1 ; b = 0,1\\ \%m_{Mg} = \dfrac{0,1.24}{5,1}.100\% = 44,44\%\ ;\ \%m_{Al} = 100\% -44,44\% = 55,56\%\\ b) n_{MgCl_2} = n_{Mg} = 0,1 \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)\\ n_{AlCl_3} = n_{Al} = 0,1 \Rightarrow m_{AlCl_3} = 0,1.133,5 = 13,35(gam)\\ c)n_{HCl} = 2n_{Mg} + 3n_{Al} = 0,5(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,5.36,5}{3,65\%} = 500(gam)\)

\(m_{dd\ sau\ pư} = 5,1 + 500 - 0,25.2 = 504,6(gam)\\ C\%_{MgCl_2} = \dfrac{9,5}{504,6}.100\% = 1,89\%\\ C\%_{AlCl_3} = \dfrac{13,35}{504,6}.100\% = 2,65\%\)

a)

Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$

$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1; b = 0,2

$m_{Zn} = 0,1.65 = 6,5(gam)$

$m_{Al} = 0,2.27 = 5,4(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$

Đáp án A

Cho Mg và Ag tác dụng với HCl chỉ có Mg phản ứng.

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.)

a) Mg + 2HCl -> MgCl2 + H2
Al + 3HCl -> AlCl3 + 3/2H2

b) Gọi a, b lần lượt là số mol Mg, Al.

nH2 = 5,6/22,4 = 0,25 (mol)

Mg + 2HCl -> MgCl2 + H2

a        2a            a          a
Al + 3HCl -> AlCl3 + 3/2H2

b           3b       b          3/2b

Ta có hệ pt: 

mhh = 24a + 27b = 5,1 (g)

nH2 = a + 3/2b = 0,25 (mol)

=> a = 0,1 (mol)

b = 0,1 (mol)

200 ml = 0,2 l

nHCl = 2a + 3b = 0,2 + 0,3 = 0,5 (mol)

=> CM ddHCl = 0,5/0,2 = 2,5 (M)

%mMg = 24a/5,1*100% = 2,4/5,1*100% = 47,06%

%mAl = 100%-47,06% = 52,94%