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- Xét A:
Giả sử \(m_{SO_2}=m_{CH_4}=16\left(g\right)\)
\(n_{SO_2}=\dfrac{16}{64}=0,25\left(mol\right);n_{CH_4}=\dfrac{16}{16}=1\left(mol\right)\)
\(\overline{M}_A=\dfrac{16+16}{0,25+1}=25,6\left(g/mol\right)\)
- Xét B:
Do \(V_{Cl_2}=V_{O_2}\Rightarrow n_{Cl_2}=n_{O_2}\)
Giả sử \(n_{Cl_2}=n_{O_2}=1\left(mol\right)\)
\(\overline{M}_B=\dfrac{1.71+1.32}{1+1}=51,5\left(g/mol\right)\)
\(d_{A/B}=\dfrac{25,6}{51,5}\approx0,497\)
a)
2Al + 2NaOH + 2H2O --> 2NaAlO2 + 3H2
2Al + 6HCl --> 2AlCl3 + 3H2
Fe + 2HCl --> FeCl2 + H2
b)\(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\) => \(n_{H_2}=0,3\left(mol\right)\)
PTHH: 2Al + 2NaOH + 2H2O --> 2NaAlO2 + 3H2
0,2<---------------------------------------0,3
=> nAl = 0,2 (mol)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2---------------------->0,3
Fe + 2HCl --> FeCl2 + H2
0,1<---------------------0,1
=> a = 0,2.27 + 0,1.56 = 11(g)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11}.100\%=49,09\%\\\%m_{Fe}=\dfrac{0,1.56}{11}.100\%=50,91\%\end{matrix}\right.\)
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\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{m}{27}\) \(\dfrac{m}{18}\) ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\dfrac{m}{65}\) \(\dfrac{m}{65}\) ( mol )
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{m}{56}\) \(\dfrac{m}{56}\) ( mol )
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(\dfrac{m}{24}\) \(\dfrac{m}{24}\) ( mol )
Ta có:
\(\dfrac{m}{18}< \dfrac{m}{24}< \dfrac{m}{56}< \dfrac{m}{65}\)
=> Al cho nhiều H2 nhất
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{Al} = \dfrac{a}{27} (mol) \Rightarrow n_{H_2} = \dfrac{3}{2}n_{Al} = \dfrac{a}{18}(mol)$
$n_{Zn} = \dfrac{b}{65}(mol) \Rightarrow n_{H_2} = n_{Zn} = \dfrac{b}{65}(mol)$
$\Rightarrow \dfrac{a}{18} = \dfrac{b}{65}$
$\Rightarrow \dfrac{a}{b} = \dfrac{18}{65}$