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\(m_{CuO}=50.20\%=10\left(g\right)\)
\(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)
\(m_{Fe_2O_3}=50-10=40\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{40}{160}=0,25\left(mol\right)\)
PTHH :
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,125 0,125 0,125
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,25 0,75 0,5
\(a,V_{H_2}=\left(0,75+0,125\right).22,4=19,6\left(l\right)\)
\(b,m_{Cu}=0,125.64=8\left(g\right)\)
\(m_{Fe}=0,5.56=28\left(g\right)\)
a.b.
\(\left\{{}\begin{matrix}n_{Fe_2O_3}=40.80\%=32g\\m_{CuO}=40-32=8g\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{32}{160}=0,2mol\\n_{CuO}=\dfrac{8}{80}=0,1mol\end{matrix}\right.\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,1 0,1 0,1 ( mol )
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,2 0,6 0,4 ( mol )
\(V_{H_2}=\left(0,1+0,6\right).22,4=15,68l\)
\(\left\{{}\begin{matrix}m_{Cu}=0,1.64=6,4g\\m_{Fe}=0,4.56=22,4g\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{6,4}{6,4+22,4}.100=22,22\%\\\%m_{Fe}=100\%-22,22\%=77,78\%\end{matrix}\right.\)
c.
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) ( Cu không phản ứng với H2SO4 loãng )
0,4 0,4 ( mol )
\(V_{H_2}=0,4.22,4=8,96l\)
\(m_{CuO}=40.20\%=8\left(g\right)\)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(n_{Fe_2O_3}=\dfrac{40-8}{160}=0,2\left(mol\right)\)
PTHH:
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,1 0,1
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,2 0,6
\(V_{H_2}=\left(0,1+0,6\right).22,4=15,68\left(l\right)\)
$m_{Fe_2O_3} = 60.80\% = 48(gam) \Rightarrow n_{Fe_2O_3} = \dfrac{48}{160} = 0,3(mol)$
$m_{CuO} = 60 - 48 = 12(gam) \Rightarrow n_{CuO} = \dfrac{12}{80} = 0,15(mol)$
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{H_2} = 3n_{Fe_2O_3} + n_{CuO} = 1,05(mol)$
$V_{H_2} = 1,05.22,4 = 23,52(lít)$
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,2 <--- 0,4
Đặt nFe2O3 = a (mol); nFeO = b (mol)
160a + 72b = 15,2 (1)
PTHH: Fe2O3 + 3H2 -> (to) 2Fe + 3H2O
Mol: a ---> 3a ---> 2a
FeO + H2 -> (to) Fe + H2O
Mol: b ---> b ---> b
2a + b = 0,2 (2)
(1)(2) => a = 0,05 (mol); b = 0,1 (mol)
mFe2O3 = 0,05 . 160 = 8 (g)
%mFe2O3 = 8/15,2 = 52,63%
%mFeO = 100% - 52,63% = 47,37%
nH2 = 0,05 . 3 + 0,1 = 0,25 (mol)
VH2 = 0,25 . 22,4 = 5,6 (l)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
a, Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20.60\%=12\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\\m_{CuO}=20-12=8\left(g\right)\Rightarrow n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\end{matrix}\right.\)
Theo pT: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,15\left(mol\right)\\n_{Cu}=n_{CuO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(m_{Cu}=0,1.64=6,4\left(g\right)\)
b, Theo PT: \(n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,325\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,325.22,4=7,28\left(l\right)\)
Bạn tham khảo nhé!
\(m_{Fe_2O_3}=\dfrac{80\cdot50}{100}=40\left(g\right)\)
\(m_{CuO}=50-40=10\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{40}{160}=0.25\left(mol\right)\)
\(n_{CuO}=\dfrac{10}{80}=0.125\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{t^0}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(n_{H_2}=3\cdot0.25+0.125=0.875\left(mol\right)\)
\(V_{H_2}=0.875\cdot22.4=19.6\left(l\right)\)
Chúc bạn học tốt <3