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\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:\(2Fe+6HCl\rightarrow2FeCl_3+3H_2\)
0,1 0,4 0,1 0,2
\(CM_{HCl}=\dfrac{n_{ct}}{V_{dd}}=\dfrac{0,4}{0,5}\)=0,8(M)
\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Theo PTHH: \(n_{H_2}=n_{Fe}=n_{H_2SO_4}=1,5\left(mol\right)\)
\(\Rightarrow m_{Fe}=1,5.56=84\left(g\right)\)
\(\Rightarrow CM_{H_2SO_4}=\dfrac{1,5}{0,5}=3M\)
Fe + H2SO4 → FeSO4 + H2
\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
a) Theo PT: \(n_{Fe}=n_{H_2}=1,5\left(mol\right)\)
\(\Rightarrow m_{Fe}=1,5\times56=84\left(g\right)\)
b) Đổi: 500 ml = 0,5 l
Theo PT: \(n_{H_2SO_4}=n_{H_2}=1,5\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{1,5}{0,5}=3\left(M\right)\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :
$n_{Fe\ pư} = n_{H_2} = \dfrac{33,6}{22,4} = 1,5(mol)$
$m_{Fe\ pư} = 1,5.56 = 84(gam)$
b)
$n_{HCl} = 2n_{H_2} = 3(mol) \Rightarrow C_{M_{HCl}} = \dfrac{3}{0,5} = 6M$
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Dễ thấy \(m_{Cu}=12,8\left(g\right)\)
Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)=n_{Fe}=n_{FeSO_4}=n_{H_2SO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{hh}=0,6\cdot56+12,8=46,4\left(g\right)\\C\%_{H_2SO_4}=\dfrac{0,6\cdot98}{196}\cdot100\%=30\%\\m_{FeSO_4}=0,6\cdot152=91,2\left(g\right)\\m_{H_2}=0,6\cdot2=1,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}-m_{Cu}=228,4\left(g\right)\)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{91,2}{228,4}\cdot100\%\approx39,93\%\)
\(Fe + H_2SO_4 \rightarrow FeSO_4 + H_2\)
Chất rắn không tan là Cu ( do Cu không pư \(H_2SO_4\))
\(n_{H_2} =\dfrac{13,44}{22,4}=0,6 mol\)
Theo PTHH
\(n_{Fe}= n_{H_2} = 0,6 mol\)
\(\Rightarrow m_{Fe} = 0,6 . 56=33,6g\)
\(m_{hh}= m_{Fe} + m_{Cu}= 33,6 + 12,8 =46,4g\)
b)
Theo PTHH
\(n_{H_2SO_4}=n_{H_2}= 0,6 mol\)
\(\Rightarrow m_{H_2SO_4}= 0,6 . 98=58,8g\)
C%\(H_2SO_4\)= \(\dfrac{58,8}{196} . 100\)%= 30%
\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
\(m_{Al}=0,2.27=5,4g\\ b.C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,45}=\dfrac{2}{3}M\\ c.2H_2+O_2\underrightarrow{t^0}2H_2O\)
0,3 0,15 0,3
\(V_{O_2}=0,15.22,4=3,36l\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ a,m_{Al}=0,2.27=5,4\left(g\right)\\ n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\\ b,C_{MddH_2SO_4}=\dfrac{0,3}{0,45}=\dfrac{2}{3}\left(M\right)\\ 2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{n_{H_2}}{2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ c,V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)
nH2 = 1,5 mol
Pt: Fe + H2SO4 --> FeSO4 + H2
....1,5 mol----------> 1,5 mol
Ta thấy: nFeSO4 = nFeSO4.10H2O = 1,5 mol
=> mFeSO4.10H2O = 1,5 . 332 = 498 (g)
FeSO4.7H2O mà