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\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)
a/ \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,3 0,6 0,3 0,3
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
Gọi a, b lần lượt là mol của Al và Zn
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a 1,5a
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b b
\(\Rightarrow\left\{{}\begin{matrix}27a+65b=9,2\\1,5a+b=\dfrac{5,6}{22,4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\)
\(\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,35\%\)
b. \(n_{H_2}=0,25mol\) \(\Rightarrow n_{HCl}=0,5mol\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25g\)
Ta có: \(10\%=\dfrac{18,25}{m_{dd}}.100\%\)
\(\Leftrightarrow m_{dd}=182,5g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow \%_{Al}=\dfrac{5,4}{10}.100\%=54\%\\ \Rightarrow \%_{Ag}=100\%-54\%=46\%\\ n_{HCl}=0,6(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,6.36,5}{10\%}=219(g)\)
$n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)$
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,2.27}{24,6}.100\%=21,95\%$
$\Rightarrow \%m_{Cu}=100-21,95=78,05\%$
$b)n_{AlCl_3}=n_{Al}=0,2(mol)$
$\Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)$
$c)n_{HCl}=3n_{Al}=0,6(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$
$a)PTHH:Fe+2HCl\to FeCl_2+H_2$
$\Rightarrow n_{Fe}=n_{H_2}=\dfrac{2,479}{24,79}=0,1(mol)$
$\Rightarrow \%m_{Fe}=\dfrac{0,1.56}{12}.100\%=46,67\%$
$\Rightarrow \%m_{Cu}=100-46,67=53,33\%$
$b)n_{FeCl_2}=n_{Fe}=0,1(mol)$
$\Rightarrow m_{FeCl_2}=0,1.127=12,7(g)$
$c)n_{HCl}=2n_{Fe}=0,2(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M$
a, Ta có: 65nZn + 27nAl = 11,9 (1)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)
⇒ mZn = 0,1.65 = 6,5 (g)
mAl = 0,2.27 = 5,4 (g)
b, Theo PT: nZnCl2 = nZn = 0,1 (mol)
nAlCl3 = nAl = 0,2 (mol)
⇒ m muối = 0,1.136 + 0,2.133,5 = 40,3 (g)
c, Theo PT: nHCl = 2nH2 = 0,8 (mol)
\(\Rightarrow m_{ddHCl}=\dfrac{0,8.36,5}{10\%}=292\left(g\right)\)
m dd sau pư = mFe + m dd HCl - mH2 thôi em nhé, Cu không phản ứng nên không cộng thêm vào.
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) Theo Pt : \(n_{H2}=n_{Fe}=n_{FeCl2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\)
\(\%m_{Cu}=100\%-56\%=44\%\)
b) Theo Pt : \(n_{H2}=2n_{HCl}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}.100\%=100\left(g\right)\)
c) \(m_{ddspu}=10+100-0,1.2=109,8\left(g\right)\)
\(C\%_{FeCl2}=\dfrac{0,1.127}{109,8}.100\%=11,57\%\)
\(a,n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,6(mol)\\ \Rightarrow m_{CT_{HCl}}=0,6.36,5=21,9(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{21,9}{28\%}=78,21(g)\\ b,n_{Mg}=n_{H_2}=0,3(mol)\\ \Rightarrow m_{Mg}=0,3.24=7,2(g)\\ \Rightarrow {\%}_{Mg}=\dfrac{7,2}{18}.100{\%}=40\%\\ \Rightarrow {\%}_{Ag}=60\%\)
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