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\(1)n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\)
Từ giả thiết và theo PT:
\(\begin{cases} 24n_{Mg}+56n_{Fe}=5,2\\ n_{Mg}+n_{Fe}=0,15 \end{cases}\\ \Rightarrow n_{Mg}=0,1(mol);n_{Fe}=0,05(mol)\)
\(\Rightarrow \begin{cases} \%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%=46,15\%\\ \%m_{Fe}=100-46,15=53,85\% \end{cases}\\ 2)\Sigma n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{1}=0,3(l)=300(ml)\)
lười làm thì đừng làm
box hóa có luật không tham khảo rồi
a)
\(n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ n_{Mg} = a\ mol; n_{Fe} = b\ mol\\ Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2 \)
Theo PTHH, ta có:
\(\left\{{}\begin{matrix}24a+56b=5,2\\a+b=0,15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
Suy ra:
\(\%m_{Mg} = \dfrac{0,1.24}{5,2}.100\% = 46,15\%\\ \%m_{Fe} = 100\% - 46,15\% = 53,85\% \)
b)
\(n_{HCl} = 2n_{H_2} = 0,15.2 = 0,3(mol)\\ \Rightarrow V_{dd\ HCl} = \dfrac{0,3}{1} = 0,3(lít) \)
Đặt :
nMg = a mol
nFe= b mol
mhh = 24a + 56b = 5.2 (g) (1)
Mg + 2HCl => MgCl2 + H2
Fe + 2HCl => FeCl2 + H2
nH2 = a + b = 0.15 (2)
(1) , (2)
a = 0.1
b = 0.05
%Mg = 2.4/5.2 * 100% = 46.15%
%Fe = 100 - 46.15 = 53.85%
nHCl = 2a + 2b = 0.05 * 2 + 0.1*2 = 0.3 (mol)
VddHCl = 0.3/1=0.3 (l)
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
\(n_{Zn} = a(mol) ; n_{Al} = b(mol) ; n_{Mg} = c(mol)\\ \Rightarrow 65a + 27b + 24c = 44,1(1)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3 H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{31,36}{22,4} = 1,4(2)\\ Mà : 2a = 3b(3)\\ (1)(2)(3) \Rightarrow a = 0,3 ; b = 0,2 ; c = 0,8\\ \%m_{Zn} = \dfrac{0,3.65}{44,1}.100\% = 44,22\%\\ \%m_{Al} = \dfrac{0,2.27}{44,1}.100\% = 12,24\%\)
\(\%m_{Mg} = 100\% -44,22\% -12,24\% = 43,54\%\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo pt: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)
\(\%m_{Fe}=100\%-19,84\%=80,16\%\)
Sửa đề : 13.9 (g)
\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(\Rightarrow m=27a+56b=13.9\left(1\right)\)
\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=1.5a+b=0.35\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Al=\dfrac{0.1\cdot27}{13.9}\cdot100\%=19.42\%\)
\(\%Fe=100-19.42=80.58\%\)