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\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)
4) x,y lần lượt là số mol của M và M2O3
=> nOxi=3y=nCO2=0,3 => y=0,1
Đề cho x=y=0,1 =>0,1M+0,1(2M+48)=21,6 =>M=56 => Fe và Fe2O3
=> m=0,1.56 + 0,1.2.56=16,8
2)X + 2HCl === XCl2 + H2
n_h2 = 0,4 => X = 9,6/0,4 = 24 (Mg)
=>V_HCl = 0,4.2/1 = 0,8 l
\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.n_{Al}=0,2\left(mol\right)\\ n_{HCl}=3n_{Al}=0,6\left(mol\right)\\ C\%_{HCl}=\dfrac{0,6.36,5}{150}.100=14,6\%\\ c.n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\\ Bảotòannguyêntố\left(H\right)\Rightarrow n_{H_2O}=n_{H_2}=0,3\left(mol\right)\\ Bảotoànkhốilượng:m_{H_2}+m_{oxit}=m_{Fe}+m_{H_2O}\\ \Rightarrow m_{Fe}=0,3.2+17,4-0,3.18=12,6\left(g\right)\\ \Rightarrow n_{Fe}=0,225\left(mol\right)\\ Tabiết:Oxitsắtlàbaogồm:Fe,O\\ \Rightarrow m_O=17,4-12,6=4,8\left(g\right)\\ \Rightarrow n_O=0,3\left(mol\right)\\ GọiCToxitsắtlà:Fe_xO_y\left(x,y>0,x,ynguyên\right)\\ Tacó:x:y=0,225:0,3=3:4\\ VậyCToxitsắtcầntìmlàFe_3O_4\)
Fe+2HCl->FeCl2+H2
0,125--0,25---0,125-0,125
m HCl=9,125 g=>n HCl=\(\dfrac{9,125}{26,5}\)=0,25 mol
=>m Fe=0,125.56=7g
=>VH2=0,125.22,4=2,8l
=>C%FeCl2=\(\dfrac{0,125.127}{7+182,5-0,25}\).100=8,388%
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)
\(b,\) Đặt \(n_{Fe}=x(mol);n_{Al}=y(mol)\)
\(\Rightarrow 56x+27y=8,3(1)\)
Theo PTHH: \(x+1,5y=0,25(2)\)
\((1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8,3}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\%\)
Bài 1
\(a.n_{H_2}=\dfrac{3,7185}{24,79}=0,15mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,3 0,1 0,15
\(\%m_{Al}=\dfrac{0,1.27}{10,7}\cdot100=25,23\%\\ \%m_{MgO}=100-25,23=74,77\%\)
\(b.n_{MgO}=\dfrac{10,7-0,1.27}{40}=0,2mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,2 0,4
\(V_{ddHCl}=\dfrac{0,3+0,4}{0,5}=1,4l\)
Bài 2:
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)
c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
Ta có: m dd sau pư = 6,5 + 100 - 0,1.2 = 106,3 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)