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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)
\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)
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:
\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
Fe+2HCl->FeCl2+H2
x---2x-----------x
Mg+2HCl->MgCl2+H2
y------2y-----------y
Ta có :
\(\left\{{}\begin{matrix}56x+24y=24\\x+y=\dfrac{13,44}{22,4}\end{matrix}\right.\)
=>x=0,3 mol, y=0,3 mol
=>%m Fe=\(\dfrac{0,3.56}{24}.100\)=70%
=>%m Mg=100-70=30%
=>VHCl=\(\dfrac{0,3.2+0,3.2}{2}\)=0,6l=600ml
b)
XCl2+2AgNO3->2AgCl+X(NO3)2
0,6--------------------1,2mol
=>m AgCl=1,2.143,5=172,2g
Bổ sung
Câu 2:
\(\left\{{}\begin{matrix}24n_{Mg}+56n_{Fe}=5,92\\95n_{Mg}+127n_{Fe}=15,86\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,06\left(mol\right)\\n_{Fe}=0,08\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,06.24=1,44\left(g\right)\\m_{Fe}=0,08.56=4,48\left(g\right)\end{matrix}\right.\)
\(2n_{Mg}+2n_{Fe}=2n_{H2}\)
\(\Rightarrow n_{H2}=0,14\left(mol\right)\)
\(n_{HCl}=2n_{H2}\Rightarrow n_{HCl}=0,28\left(mol\right)\)
\(\Rightarrow V_{Dd\left(HCl\right)}=\frac{0,28}{2}=0,14\left(l\right)=140\left(ml\right)\)
Gọi nFe = nAl = a mol
có mX = mAl + mFe = 27a +56a =83a = 7,47 gam
⇒ a =0,09 mol
PTHH :
2Al + 6HCl ----> 2AlCl3 + 3H2
Fe + 2HCl -----> FeCl2 + H2
Theo PT trên nH2SO4 = 3nAl + 2nFe = 3.0,09 +2.0,09 =0,45 mol
\(=>C\%HCL=\frac{0,45.36,5}{400}.100\%=4,11\%\)
có nAlCl3 = nAl = 0,09 mol
nFeCl2 = nFe =0,2 mol
⇒ m muối khan = mAlCl3 + mFeCl2 = 0,09.133,5 + 0,09.127 = 23,445 gam