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a) Coi X là kim loại R hóa trị n
\(2R + 2nHCl \to 2RCl_n + nH_2\\ n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ \Rightarrow n_R = \dfrac{2}{n}n_{H_2} = \dfrac{0,3}{n}(mol)\\ 2R + 2nH_2O \to 2R(OH)_n + nH_2\\ n_{H_2O} = \dfrac{10,8}{18} = 0,6(mol)\\ \Rightarrow n_R = \dfrac{1}{n}n_{H_2O} = \dfrac{0,6}{n}(mol)\\ \)
Suy ra: \(\dfrac{m_1}{m_2} = \dfrac{0,3}{n} : \dfrac{0,6}{n} = \dfrac{1}{2}\)
b)
\(m_2 =2m_1 \\ \Rightarrow C_{M_{HCl\ TN_2}} = 2C_{M_{HCl\ TN_1}} = 0,5.2 = 1M\)
\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
b) \(n_{Fe}=n_{H2}=n_{H2SO4}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Al2O3}=15,8-5,6=10,2\left(g\right)\)
c) Ta có : \(n_{Al2O3}=\dfrac{10,2}{102}=0,1\left(mol\right)\Rightarrow n_{H2SO4}=3n_{Al2O3}=0,3\left(mol\right)\)
\(C_{MddH2SO4}=\dfrac{0,1+0,3}{0,2}=2M\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)
Bài 1
\(a)n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
\(\%m_{Fe}=\dfrac{0,15.56}{12}\cdot100\%=70\%\\ \%m_{FeO}=100\%-70\%=30\%\\ b)n_{FeO}=\dfrac{12-0,15.56}{72}=0,05mol\\ FeO+2HCl\rightarrow FeCl_2+H_2O\)
0,05 0,1
\(V_{ddHCl}=\dfrac{0,1+0,3}{2}=0,2l\)
Bài 2
\(a)Na_2O+H_2O\rightarrow2NaOH\\ b)BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ c)BaSO_4?\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4+Cu\left(OH\right)_2\\ d)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ e)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\)
a)\(n_{HCl}=\dfrac{10,65}{36,5}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
b)\(V_{H_2}=0,15\cdot22,4=3,36l\)
c)\(n_{CuO}=\dfrac{16}{80}=0,2mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,2 0,15 0,15
\(m_{Cu}=0,15\cdot64=9,6g\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$
b)
Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$
$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$
c) $Fe + CuSO_4 \to FeSO_4 + Cu$
$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$
$= 3,6 + 0,15.64 = 13,2(gam)$
a)
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$
Theo PTHH : $n_{HCl} = 3n_{Al} = 1,2(mol)$
$\Rightarrow m = \dfrac{1,2.36,5}{14,6\%} = 300(gam)$
b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$M_xO_y + yH_2 \xrightarrow{t^o}xM + yH_2O$
Theo PTHH : $n_{oxit} = \dfrac{1}{y}.n_{H_2} = \dfrac{0,6}{y}(mol)$
$\Rightarrow \dfrac{0,6}{y}(Mx + 16y) = 34,8$
$\Rightarrow \dfrac{x}{y}.M = 42$
Với x = 3 ; y = 4 thì $M = 56(Fe)$
Vậy oxit là $Fe_3O_4$
\(a) 2Na + 2HCl \to 2NaCl + H_2\\ Ba + 2HCl \to BaCl_2 + H_2\\ 2Na + 2H_2O \to 2NaOH + H_2\\ Ba + 2H_2O \to Ba(OH)_2 + H_2\)
\(TN1 : n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ n_{Na} = x ; n_{Ba} = y\\ n_{H_2} = 0,5x + y = 0,15\\ TN2 : n_{Na} = xk ; n_{Ba} = yk\\ n_{H_2O} = n_{Na} + 2n_{Ba} =xk + 2yk = k.0,15.2 = \dfrac{10,8}{22,4} = 0,45\\ \Rightarrow k = 1,5\\ Suy\ ra: \dfrac{a}{b} = k = 1,5\)