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a)
$2K + 2H_2O \to 2KOH + H_2$
$BaO + H_2O \to Ba(OH)_2$
Theo PTHH :
$n_K = 2n_{H_2} = 0,2(mol)$
$\%m_K = \dfrac{0,2.39}{23,1}.100\% = 33,77\%$
$\%m_{BaO} = 100\%- 33,77\% = 66,23\%$
b)
$n_{BaO} = \dfrac{23,1 - 0,2.39}{153} = 0,1(mol)$
$m_{dd} = 23,1 + 177,1 - 0,1.2 = 200(gam)$
$C\%_{KOH} = \dfrac{0,2.56}{200}.100\% = 5,6\%$
$C\%_{Ba(OH)_2} = \dfrac{0,1.171}{200}.100\% = 8,55\%$
c)
$KOH + HCl \to KCl + H_2O$
$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} + n_{KOH} = 0,4(mol)$
$V = \dfrac{0,4}{0,5} = 0,8(lít) = 800(ml)$
a, Ta có: 27nAl + 56nFe = 22 (1)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)
b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)
a, Ta có: 27nAl + 56nFe = 27,8 (1)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{17,353}{24,79}=0,7\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{27,8}.100\%\approx19,42\%\\\%m_{Fe}\approx80,58\%\end{matrix}\right.\)
b, \(n_{H_2SO_4}=n_{H_2}=0,7\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,7}{0,5}=1,4\left(M\right)\)
a) Gọi $n_{Al} =a (mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 13(1)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{6,72}{22,4} = 0,3(2)$
Từ (1)(2) suy ra : $a = \dfrac{1}{15} ; b = 0,2$
$\%m_{Al} = \dfrac{ \dfrac{1}{15}.27}{13}.100\% = 13,8\%$
$\%m_{Fe} = 100\% - 13,8\% = 86,2\%$
b) $n_{HCl} = 2n_{H_2} = 0,3.2 = 0,6(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,6}{0,15} = 4M$
c) $n_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13 + 0,6.36,5 - 0,3.2 = 34,3(gam)$
\(n_{Fe}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,01 0,02 0,01
a) \(n_{Fe}=\dfrac{0,01.1}{1}=0,01\left(mol\right)\)
\(m_{Fe}=0,01.56=0,56\left(g\right)\)
\(m_{Cu}=1,2-0,56=0,64\left(g\right)\)
0/0Fe = \(\dfrac{0,56.100}{1,2}=46,67\)0/0
0/0Cu = \(\dfrac{0,64.100}{1,2}=53,33\)0/0
b) \(n_{HCl}=\dfrac{0,01.2}{1}=0,02\left(mol\right)\)
⇒ \(m_{HCl}=0,02.36,5=0,73\left(g\right)\)
\(C_{ddHCl}=\dfrac{0,73.100}{10}=7,3\)0/0
Chúc bạn học tốt
a, gọi a= nFe
b= nFeO
=> 56a + 72b= 12,8 (1)
Fe +H2SO4 -> FeSO4 +H2
a b b a
FeO +H2SO4 -> FeSO4 +H2O
b b b
a=nH2 = 2,24/22,4= 0,1 mol
từ (1) => b= 0,1
mFe= 56.0,1=5,6(g)
m FeO = 72.0,1= 7,2(g)
b, nH2SO4 (bđ) = 0,25 mol
nH2SO4 pứ = a+b =0,2 mol
=> nH2SO4 dư = 0,25-0,2=0,05 mol
2NaOH +H2SO4 -> Na2SO4 +2H2O
0,1 0,05
V(NaOH)= 0,1/ 1= 0,1 lit =100ml
\(a,n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ b,n_{H_2}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ c,n_{FeCl_2}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,93\%\)
