Cho 12,8g hỗn hợp Fe và FeO tác dụng với dung dịch HCl 2M thì thu được 2,24l khí(đktc).Hãy tính.a/Khối lượng mỗi chất trong hỗn hợp.b/Thành phần % khối lượng mỗi chất trong hỗn .c/Thể tích dung dịch axit đã dùng.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Fe + 2HCl -> FeCl2 + H2
0.2 0.1
FeO + 2HCl -> FeCl2 + H2O
0.1 0.2
a.\(nH2=\dfrac{2.24}{22.4}=0.1mol\)
\(\%mFe=\dfrac{0.1\times56\times100}{12.8}=43.8\%\)
\(\%mFeO=100-43.8=56.2\%\)
b.\(nFeO=\dfrac{12.8-\left(0.1\times56\right)}{56+16}=0.1mol\)
\(V_{HCl}=\dfrac{0.2+0.2}{2}=0.2l\)
\(a/n_{khí}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Fe}=a;n_{FeCO_3}=b\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCO_3+2HCl\rightarrow FeCl_2+CO_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}56a+116b=7,2\\a+b=0,1\end{matrix}\right.\\ \Rightarrow a=\dfrac{11}{150};b=\dfrac{2}{75}\\ \%m_{Fe}=\dfrac{11:150.56}{7,2}\cdot100\%=57,04\%\\ \%m_{FeCO_3}=100\%-57,04\%=42,96\%\\ b/n_{FeCl_2}=\dfrac{11}{150}+\dfrac{2}{75}=0,1mol\\ C_{\%FeCl_2}=\dfrac{0,1.127}{7,2+94,9-\dfrac{11}{150}\cdot2-\dfrac{2}{75}\cdot44}\cdot100\%=12,6\%\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$
b)
Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$
$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$
c) $Fe + CuSO_4 \to FeSO_4 + Cu$
$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$
$= 3,6 + 0,15.64 = 13,2(gam)$
\(n_{Fe}=x(mol);n_{Mg}=y(mol)\\ \Rightarrow 56x+24y=10-2=8(1)\\ Fe+2HCl\to FeCl_2+H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow x+y=\dfrac{4,48}{22,4}=0,2(2)\\ (1)(2)\Rightarrow x=y=0,1(mol)\\ a,\begin{cases} \%_{Fe}=\dfrac{56.0,1}{10}.100\%=56\%\\ \%_{Mg}=\dfrac{24.0,1}{10}.100\%=24\%\\ \%_{Cu}=\dfrac{2}{10}.100\%=20\% \end{cases}\\ \)
\(b,\Sigma n_{HCl}=2(x+y)=0,4(mol)\\ \Rightarrow V=\dfrac{0,4}{2}=0,2(l)\)
nH2=0,1 mol
Fe + 2HCl → FeCl2 + H2
0,1 mol 0,1 mol
mFe=0,1.56=5,6g
%mFe=\(\frac{5,6}{26}\).100=21,5%
m(feo và fe3o4)=26-5,6=20,4g
ta có phương trình cho nhận electron
Fe+2 ---------> Fe+3 + 1e N+5 + 3e-----------> N+2
x mol x mol 0,45 mol 0,15 mol
Fe3+8/3 ------------> 3Fe+3 + 1e
y mol y mol
Fe0 ------------> Fe+3 + 3e
0,1 mol 0,3mol
ta có hệ phương trình
\(\begin{cases}72x+232y+5,6=26\\0,3+x+y=0,45\end{cases}\) =>\(\begin{cases}x=0,09\\y=0,06\end{cases}\)
mFeO=0,09.72=6,48 g %mFeO=\(\frac{6,48}{26}.100=24,9\%\)
mFe3O4=13,92g %mFe3O4= 100%-(21,5%+ 24,9%)=53,6%
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+65y=21,4\\x+y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4g\)
\(\Rightarrow m_{Zn}=0,2.65=13g\)
\(\%m_{Fe}=\dfrac{8,4}{21,4}.100=39,25\%\)
\(\%m_{Zn}=100\%-39,25\%=60,75\%\)
\(m_{FeCl_2}=0,15.127=19,05g\)
\(m_{ZnCl_2}=0,2.136=27,2g\)
\(a,\) Đặt \(\begin{cases} n_{Fe}=x(mol)\\ n_{Al}=y(mol) \end{cases}\Rightarrow 56x+27y=22(1)\)
\(n_{H_2}=\dfrac{17,92}{22,4}=0,8(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow x+1,5y=0,8(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,2(mol)\\ y=0,4(mol) \end{cases} \Rightarrow \begin{cases} \%_{Fe}=\dfrac{0,2.56}{22}.100\%=50,91\%\\ \%_{Al}=100\%-50,91\%=49,09\% \end{cases} \)
\(b,\Sigma n_{HCl}=2n_{Fe}+3n_{Al}=0,4+1,2=1,6(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1,6.36,5}{3,7\%}=1578,38\%\)
Bài 1
\(a)n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
\(\%m_{Fe}=\dfrac{0,15.56}{12}\cdot100\%=70\%\\ \%m_{FeO}=100\%-70\%=30\%\\ b)n_{FeO}=\dfrac{12-0,15.56}{72}=0,05mol\\ FeO+2HCl\rightarrow FeCl_2+H_2O\)
0,05 0,1
\(V_{ddHCl}=\dfrac{0,1+0,3}{2}=0,2l\)
Bài 2
\(a)Na_2O+H_2O\rightarrow2NaOH\\ b)BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ c)BaSO_4?\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4+Cu\left(OH\right)_2\\ d)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ e)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\)
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
Fe+2HCl->FeCl2+H2
0,1-------0,2---------------0,1
FeO+2HCl->FeCl2+H2O
0,1----0,2---------------0,1
Ta có :
n H2=\(\dfrac{2,24}{22,4}=0,1mol\)
mFe=0,1.56=5,6g
=>m FeO=7,2g ->n FeO=\(\dfrac{7,2}{72}=0,1mol\)
b) %mFe=\(\dfrac{5,6}{12,8}.100=43,75\%\)
=>%mFeO=56,25%
=>VHCl=\(\dfrac{0,2}{2}=0,1l=100ml\)