\(n_{H2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

         1         2              1           1

         a       0,4           0,2           1a

         \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

           1          2            1          1

          b         0,3         0,15        1b

a) Gọi a là số mol của Mg

           b là số mol của Fe

\(m_{Mg}+m_{Fe}=13,2\left(g\right)\)

⇒ \(n_{Mg}.M_{Mg}+n_{Fe}.M_{Fe}=13,2g\)

 ⇒ 24a + 56b = 13,2g (1)

Theo phương trình : 1a + 1b = 0,35(2)

Từ(1),(2), ta có hệ phương trình : 

            24a + 56b = 13,2g

              1a + 1b = 0,35

               ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\)

\(m_{Mg}=0,2.24=4,8\left(g\right)\)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

⁰/₀Mg = \(\dfrac{4,8.100}{13,2}=36,36\)⁰/₀

⁰/₀Fe = \(\dfrac{8,4.100}{13,2}=63,64\)⁰/₀

b) \(n_{HCl\left(tổng\right)}=0,4+0,3=0,7\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddHCl}}=\dfrac{0,7}{0,2}=3,5\left(M\right)\)

c) \(m_{muối.clorua}=\left(0,2.95\right)+\left(0,15.127\right)=38,05\left(g\right)\)

 Chúc bạn học tốt

C là tính tổng khối lượng nha

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: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)

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

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

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

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)

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

Gọi \(n_{Fe}=x\left(mol\right)\)\(;n_{Zn}=y\left(mol\right)\)

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

Ta có:  \(\left\{{}\begin{matrix}56x+65y=18,6\\2x+2y=2n_{H_2}=0,6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,1\cdot56}{18,6}\cdot100\%=30,11\%\)

\(\%m_{Zn}=100\%-30,11\%=69,89\%\)

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

0,1     0,2

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2     0,4

\(n_{HCl}=0,2+0,4=0,6mol\)

\(C_M=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3M\)

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

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=0,1(mol)\\ a,\%_{Zn}=\dfrac{0,1.65}{9,7}.100\%=67,01\%\\ \Rightarrow \%_{Cu}=100\%-67,01\%=32,99\%\\ b,n_{HCl}=2n_{H_2}=0,2(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{50}.100\%=14,6\%\\ c,PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{H_2}=0,1(mol)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5(l)\)