1.

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

\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{Zn}=n_{H2}=0,25\left(mol\right)\)

\(\rightarrow,m_{Zn}=0,25.65=16,25\left(g\right);m_{Cu}=30-16,25=13,75\left(g\right)\)

b)

\(\%m_{Zn}=\frac{16,25}{30}.100\%=54,17\%\)

\(\%m_{Cu}=100\%-54,17\%=45,83\%\)

c)

\(n_{HCl}=2n_{H2}=0,5\left(mol\right)\)

\(C\%_{HCl}=\frac{0,5.36,5}{200}.100\%=9,125\%\)

2.

a)

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

\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)\(m_{Fe}=0,15.65=8,4\left(g\right),m_{Ag}=15-8,4=6,6\left(g\right)\)

b)

\(\%m_{Fe}=\frac{8,4}{15}.100\%=56\%\)

\(\%m_{Ag}=100\%-56\%=44\%\)

c)

\(n_{HCl}=2n_{H2}=0,3\left(mol\right)\)

\(\rightarrow m_{dd_{HCL}}=\frac{0,3.36,5}{15,6\%}=70,19\left(g\right)\)

a, Ta có: 24n_Mg + 56n_Fe = 9,2 (g) (1)

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

BT e, có: 2n_Mg + 2n_Fe = 2n_H2 = 0,5 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\\\%m_{Fe}\approx60,87\%\end{matrix}\right.\)

b, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

Refer

em ê

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giỏi quá pé oiiii

1) Ptpư:

2Al   +  6HCl \(\rightarrow\) 2AlCl₃    +  3H₂

Fe    + 2HCl \(\rightarrow\) FeCl₂   + H₂

Cu   +  HCl \(\rightarrow\) không phản ứng

=> 0,6 gam chất rắn còn lại chính là Cu:

Gọi x, y lần lượt là số mol Al, Fe

Ta có:

3x + 2y = 2.0,06 = 0,12

27x + 56 y = 2,25 – 0,6 = 1,65

=> x = 0,03 (mol) ; y = 0,015 (mol)

=> \(\%Cu=\frac{0,6}{2,25}.100\%=26,67\%\); \(\%Fe=\frac{56.0,015}{2,25}.100\%=37,33\%\); %Al = 36%

2) \(n_{SO_2}=\frac{1,344}{22,4}=0,06mol\); m _{(dd KOH)} = 13,95.1,147 = 16 (gam)

=> m_KOH = 0,28.16 = 4,48 (gam)=> n_KOH = 0,08 (mol)=> \(1<\)\(\frac{n_{KOH}}{n_{SO_2}}<2\)

=> tạo ra hỗn hợp 2 muối: KHSO₃:  0,04 (mol)  và K₂SO₃:  0,02 (mol)

Khối lượng dung dịch sau pu = 16 + 0,06.64 = 19,84 gam

=> \(C\%\left(KHSO_3\right)=\frac{0,04.120}{19,84}.100\%\)\(=24,19\%\)

\(C\%\left(K_2SO_3\right)=\frac{0,02.158}{19,84}.100\%\)\(=15,93\%\)

bạn chỉ mình tại sao 3X+2Y=0,12 đc ko

Câu 1:

Gọi số mol Al là x; Zn là y

\(\rightarrow27x+65y=18,4\)

\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)

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

\(\rightarrow n_{H2}=1,5n_{Al}+n_{Zn}=1,5x+y=\frac{1}{2}=0,5\left(mol\right)\)

Giải được: \(x=y=0,2\)

\(\Rightarrow m_{Al}=27x=5,4\left(g\right)\Rightarrow\%m_{Al}=\frac{5,4}{18,4}=29,3\%\Rightarrow\%m_{Zn}=70,7\%\)Câu 2:

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

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

Ta có: \(n_{H2}=n_{Fe}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

Muối thu được là FeCl2

\(\rightarrow n_{FeCl2}=\frac{38,1}{56+35,5.2}=0,3\left(mol\right)\)

Ta có: \(n_{FeCl2}=n_{Fe}+n_{FeO}\rightarrow n_{FeO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{FeO}=0,2.\left(56+16\right)=14,4\left(g\right)\)

Câu 3 :

Cu không tác dụng với HCl, chỉ có Zn phản ứng.

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

Ta có: \(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo phản ứng: \(n_{Zn}=n_{H2}=0,2\left(mol\right)\rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(\rightarrow\%m_{Zn}=\frac{13}{20}=65\%\rightarrow\%m_{Cu}=35\%\)

Ta có: \(n_{HCl}=2n_{H2}=0,2.2=0,4\left(mol\right)\)

\(\Rightarrow V_{HCl}=\frac{0,4}{2}=0,2\left(l\right)\)

Câu 4:

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

\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)

Gọi số mol Fe là x; Al là y

\(\rightarrow56x+27y=22\)

Ta có: \(n_{H2}=n_{Fe}=1,5n_{Al}=x+1,5y=\frac{17,92}{22,4}=0,8\left(mol\right)\)

Giải được: \(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)

\(\rightarrow\%m_{Fe}=\frac{11,2}{22}=50,9\%\rightarrow\%m_{Al}=49,1\%\)

Ta có: \(n_{HCl}=2n_{H2}=1,6\left(mol\right)\)

\(\rightarrow m_{HCl}=1,6.36,5=58,4\left(g\right)\)

\(\rightarrow m_{dd_{HCl}}=\frac{58,4}{7,3\%}=800\left(g\right)\)

Câu 5:

Gọi chung 2 kim loại là R hóa trị I

\(R+HCl\rightarrow RCl+\frac{1}{2}H_2\)

Ta có: \(n_{H2}=\frac{0,448}{22,4}=0,02\left(mol\right)\rightarrow n_{RCl}=2n_{H2}=0,04\left(mol\right)\)

\(\rightarrow m_{RCl}=0,04.\left(R+35,5\right)=2,58\rightarrow R=29\)

Vì 2 kim loại liên tiếp nhau \(\rightarrow\) 2 kim loại là Na x mol và K y mol

\(\rightarrow x+y=n_{RCl}=0,04\left(mol\right)\)

\(m_{hh}=m_R=23x+39y=0,04.29=1,16\left(g\right)\)

Giải được: \(\rightarrow\left\{{}\begin{matrix}x=0,025\\y=0,015\end{matrix}\right.\)

\(\rightarrow m_{Na}=0,575\left(g\right)\)

\(\rightarrow\%m_{Na}=\frac{0,575}{1,16}=49,57\%\rightarrow\%m_K=50,43\%\)

Câu 6:

Khối lượng mỗi phần là 35/2=17,5g

Gọi số mol Fe, Cu, Al là a, b, c

Ta có \(56a+64b=27c=17,5\)

Phần 1: \(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

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

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\Rightarrow a=1,5b=n_{H2}=0,3\)

Phần 2: \(n_{Cl2}=\frac{10,64}{22,4}=0,475\left(mol\right)\)

\(2Fe+3Cl_2\rightarrow2FeCl_3\)

\(Cu+Cl_2\rightarrow CuCl_2\)

\(2Al+3Cl_2\rightarrow2AlCl_3\)

\(\Rightarrow1,5a+b+1,5c=n_{Cl2}=0,465\)

\(\rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\\c=0,1\end{matrix}\right.\)

\(\rightarrow\%m_{Fe}=\frac{0,15.56}{17,5}=48\%\)

\(\rightarrow\%m_{Cu}=\frac{0,1.64}{17,5}=36,57\%\)

\(\rightarrow\%m_{Al}=100\%-48\%-36,57\%=15,43\%\)

Câu 1

2Al+6HCl--->2Alcl3+3H2

x-----------------------1,5x

Zn+2HCl---->Zncl2+H2

y---------------------------y

n H2=1/2=0,5(mol)

Theo bài ta có hpt

\(\left\{{}\begin{matrix}27x+65y=18,4\\1,5x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

%m Al=0,2.27/18,4.100%=29,35%

%m Zn=100%-29,35=70,65%

Câu 2.

Fe+2HCl---->FeCl2+H2

FeO+2HCl--->FeCl2+H2

n H2=2,24/22,4=0,1(mol)

m H2=0,2(g)

n Fe=n H2=0,2(mol)

m Fe=0,2.56=11,2(g)

n FeCl2(1)=2n H2=0,2(mol)

m FeCl2(1)=0,2.127=25,4(g)

m FeCl2(PT2)=38,1-25,4=12,7(g)

n FeCl2=12,7/127=0,1(mol)

n FeO=n FeCl2=0,1(mol)

m FeO=0,1.72=7,2(g)

3.

Zn+2HCl--->ZnCl2+H2

n H2=4,48/22,4=0,2(mol)

n Zn=n H2=0,2(mol)

m Zn=0,2.56=11,2(g)

%m Zn=11,2/20.100%=56%

%m Cu=100-56=34%

b) n HCl=2n H2=0,4(mol)

V H2=0,4/2=0,2(l)

4.

a) Fe+2HCl---.FeCl2+H2

x-----------------------------x(mol)

2Al+6HCl--->AlCl3+3H2

y------------------------------1,5y

n H2=17,92/22,4=0,89mol)

Theo bài ta có hpt

\(\left\{{}\begin{matrix}56x+27y=22\\x+1,5y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)

%m Fe=0,2.56/22.100%=50,9%

%m Al=100-50,9=49,1%

b) n HCl=2n H2=1,6(mol)

m HCl=1,6.36,5=58,4(g)

m dd HCl=58,4.100/7,3=800(g)