\(2Al+6HCl-->2AlCl3+3H2\)

\(Fe+2HCl-->FeCl+H2\)

\(Zn+2HCl-->ZnCl2+H2\)

\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{H2}=0,2.2=0,4\left(g\right)\)

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

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{muối}=m_{KL}+m_{HCl}-m_{H2}=10+14,6-0,4=24,2\left(g\right)\)

Bài 1:

\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

PTHH: Fe + 2HCl → FeCl₂ + H₂

\(m_{H_2}=0,16.2=0,32\left(g\right)\)

\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)

Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)

Bài 12:

Theo ĐLBTKL, ta có:

\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

Zn+2HCl→ZnCl2+H2

 2Al+6HCl→2AlCl3+3H2

Gọi số mol HCllà x

n H2=\(\dfrac{1}{2}\)n HCl=0,5x

Bảo toàn khối lượng: mkl+mHCl=mmuối+mH2

6,05+36,5x=13,15+0,5x.2

→x=0,2 mol

mHCl=0,2.36,5=7,3 gam

=>m =mddHCl=\(\dfrac{7,3}{10\%}\)=73gam

a.\(Fe+2HCl->FeCl_2+H_2\)

b.\(FeO+2HCl->FeCl_2+H_2O\)

\(n_{H_2}=0,2\left(mol\right)\)

\(\%mFe=\dfrac{0,2.56}{20}.100\)

\(\%mFe=56\%\)

\(=>\%mFeO=100-56=44\%\)

= 22,2 + 38,325 – 1,05 = 59,475 (gam).

Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\) => 65a + 56b + 27c = 10,65 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            Fe + 2HCl --> FeCl₂ + H₂

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

=> \(n_{H_2}=a+b+1,5c=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) (2)

PTHH: Zn + Cl₂ --t^o--> ZnCl₂

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

            2Al + 3Cl₂ --t^o--> 2AlCl₃

=> \(n_{Cl_2}=a+1,5b+1,5c=\dfrac{5,6}{22,4}=0,25\left(mol\right)\) (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\\c=0,05\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Fe}=0,05.56=2,8\left(g\right)\\m_{Al}=0,05.27=1,35\left(g\right)\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{6,5}{10,65}.100\%=61,033\%\\\%m_{Fe}=\dfrac{2,8}{10,65}.100\%=26,291\%\\\%m_{Al}=\dfrac{1,35}{10,65}.100\%=12,676\%\end{matrix}\right.\)

b) n_HCl = 2a + 2b + 3c = 0,45 (mol)

=> m_HCl = 0,45.36,5 = 16,425 (g)

=> \(a\%=C\%=\dfrac{16,425}{200}.100\%=8,2125\%\)

c) m_{dd sau pư} = 10,65 + 200 - 0,225.2 = 210,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{210,2}.100\%=6,47\%\\C\%_{FeCl_2}=\dfrac{0,05.127}{210,2}.100\%=3,02\%\\C\%_{AlCl_3}=\dfrac{0,05.133,5}{210,2}.100\%=3,176\%\end{matrix}\right.\)

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)

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