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a)\(Mg+2HCl--.MgCl2+H2\)
X------------------------------------x(mol)
\(2Al+6HCl-->2AlCl3+3H2\)
y--------------------------------------1,5y(mol)
\(Zn+2HCl-->ZnCl2+H2\)
z-------------------------------------z(mol)
\(n_{H2}=\frac{16,352}{22,4}=0,73\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+27y+65z=19,46\\24x-27y=0\\x+1,5y+z=0,73\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,27\\y=0,24\\z=0,1\end{matrix}\right.\)
\(\%m_{Mg}=\frac{0,27.24}{19,46}.100\%=33,3\%\)
\(\%m_{Al}=\frac{0,24.27}{19,46}.100\%=33,3\%\)
\(\%m_{Zn}=100-33,3-33,3=33,4\%\)
b)\(n_{HCl}=2n_{H2}=1,46\left(mol\right)\)
\(V_{HCl}=\frac{1,46}{2}=0,73\left(l\right)\)
\(n_{H_2}=\dfrac{4,368}{22,4}=0,195mol\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+24y=3,87\left(1\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(\Rightarrow1,5x+y=0,195\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,09\\y=0,06\end{matrix}\right.\)
\(m_{Al}=0,09\cdot27=2,43g\)
\(m_{Mg}=0,06\cdot24=1,44g\)
a:
Cu không tác dụng với HCl
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(m_{Mg}=0.2\cdot24=4.8\left(g\right)\)
\(\%Mg=\dfrac{4.8}{10}=48\%\)
b: \(m_{MgCl_2}=0.2\left(24+35.5\cdot2\right)=19\left(g\right)\)
\(m_{dd\left(Saupư\right)}=4.8+90-0.2\cdot2=94.4\)
=>\(C\%=\dfrac{19}{94.4}\simeq20,13\%\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
x 2x x x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y 2y y y
\(\left\{{}\begin{matrix}x+y=0,5\\24x+56y=23,2\end{matrix}\right.\)
\(\Leftrightarrow x=0,15;y=0,35\)
\(a,m_{Mg}=0,15.24=3,6\left(g\right)\)
\(m_{Fe}=19,6\left(g\right)\)
\(b,m_{HCl}=\left(0,3+0,7\right).36,5=36,5\left(g\right)\)
\(m_{ddHCl}=1,14.200=228\left(g\right)\)
\(C\%=\dfrac{36,5}{228}.100\%=16\%\)
\(a.n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_{Mg}=a;n_{Fe}=b\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+56b=23,2\\a+b=0,5\end{matrix}\right.\\ \Rightarrow a=0,15;b=0,35mol\\ m_{Mg}=0,15.24=3,6g\\ m_{Fe}=23,2-3,6=19,6g\\ b.m_{HCl}=\left(0,15+0,35\right).2.36,5=36,5g\\ m_{ddHCl}=1,14.200=228g\\ C_{\%HCl}=\dfrac{36,5}{228}\cdot100=16,01\%\)
nH2 = 0.73
gọi nMg=x mol ,nAl=y mol,nZn = z mol
Có hệ sau :\(\begin{cases}24x+27y+65z=19,46\\24x=27y\\2x+y+2z=0,73.2\end{cases}\)
=>x = 0.27 mol,
y = 0.24mol
,z = 0.1 mol
mMg = mAl = 6.48 => % Al = %Mg = 33.3%
=> % Zn = 33.4%
Cảm ơn