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Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 27y = 7,8 (1)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
BT e, có: 2x + 3y = 0,8 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{7,8}.100\%\approx30,77\%\\\%m_{Al}\approx69,23\%\end{matrix}\right.\)
b, BTNT Mg và Al, có:
nMgCl2 = nMg = 0,1 (mol)
nAlCl3 = nAl = 0,2 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,1.95}{0,1.95+0,2.133,5}.100\%\approx26,24\%\\\%m_{AlCl_3}\approx73,76\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\\n_{Fe}=c\left(mol\right)\end{matrix}\right.\)⇒ 24a + 27b + 56c = 26,05(1)
\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al +6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\)
\(Mg + Cl_2 \xrightarrow{t^o} MgCl_2\\ 2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Cl_2} = a + 1,5b + 1,5c = \dfrac{17,36}{22,4} = 0,775(3)\)
Từ (1)(2)(3) suy ra: a = 0,325 ; b = -0,05 ; c = 0,35
→ Sai đề.
\(a)n_{Mg} = a ; n_{Al} = b \Rightarrow 24a +27b = 5,1(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{5,6}{22,4} = 0,25(2)\\ (1)(2) \Rightarrow a = 0,1 ; b = 0,1\\ \%m_{Mg} = \dfrac{0,1.24}{5,1}.100\% = 44,44\%\ ;\ \%m_{Al} = 100\% -44,44\% = 55,56\%\\ b) n_{MgCl_2} = n_{Mg} = 0,1 \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)\\ n_{AlCl_3} = n_{Al} = 0,1 \Rightarrow m_{AlCl_3} = 0,1.133,5 = 13,35(gam)\\ c)n_{HCl} = 2n_{Mg} + 3n_{Al} = 0,5(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,5.36,5}{3,65\%} = 500(gam)\)
\(m_{dd\ sau\ pư} = 5,1 + 500 - 0,25.2 = 504,6(gam)\\ C\%_{MgCl_2} = \dfrac{9,5}{504,6}.100\% = 1,89\%\\ C\%_{AlCl_3} = \dfrac{13,35}{504,6}.100\% = 2,65\%\)
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
1.
a, \(2Fe+3Cl_2\underrightarrow{^{to}}2FeCl_3\)
\(m_{FeCl_3}=\frac{16,25.100}{100}=16,25\left(g\right)\)
b, \(n_{FeCl_3}=\frac{16,25}{162,5}=0,1\left(mol\right)\)
\(\rightarrow n_{Fe}=n_{FeCl_3}=0,1\left(mol\right)\)
\(\rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(n_{CL2}=\frac{3}{2}n_{FeCl3}=0,15\left(mol\right)\)
\(\rightarrow V_{CL2}=0,15.22,4=3,36\left(l\right)\)
Bài 2 :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(\rightarrow n_{Zn}=n_{H2}=0,1\left(mol\right)\)
\(m_{Zn}=0,1.65=6,5\left(g\right),m_{ZnO}=10,55-6,5=4,05\left(g\right)\)
b)
\(n_{ZnO}=\frac{4,05}{81}=0,05\left(mol\right)\)
\(n_{HCl}=0,05.2+0,1.2=0,3\left(mol\right)\)
\(\rightarrow m_{dd_{HCl}}=\frac{0,3.36,5}{10\%}=109,5\left(g\right)\)
Bài 3 : Xem lại đề
Bài 4:
a)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCl}=0,25.2=0,5\left(mol\right)\)
Gọi a là số mol Fe b là số mol Zn
Giải hệ phương trình :
\(\left\{{}\begin{matrix}56a+65b=14,9\\2a+2b=0,5\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,15.56}{14,9}.100\%=56,38\%,\%m_{Zn}=100\%-56,38\%=43,62\%\)
b)
\(n_{H2}=\frac{n_{HCl}}{2}=\frac{0,5}{2}=0,25\left(mol\right)\)
\(\rightarrow V_{H2}=0,25.22,4=5,6\left(l\right)\)
Bài 5 :
m tăng thêm=mKl-mH2
\(\rightarrow m_{H2}=7,8-7=0,8\left(g\right)\)
\(\rightarrow n_{H2}=\frac{0,8}{2}=0,4\left(mol\right)\)
Gọi a là số mol Al b là số mol Mg
Giải hệ phương trình :
\(\left\{{}\begin{matrix}27a+24b=7,8\\1,5a+b=0,4\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right),m_{Mg}=0,1.24=2,4\left(g\right)\)