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\(\left\{{}\begin{matrix}CuO:a\\Fe2O3:2a\end{matrix}\right.\)
a.\(80a+320a=24\Leftrightarrow a=0.06\)
\(\Rightarrow\left\{{}\begin{matrix}CuO=0.06\\Fe2O3=0.12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}CuO=4.8g\\Fe2O3=19.2g\end{matrix}\right.\)
b.\(CuO+H2\rightarrow Cu+H2O\)
a a a
\(Fe2O3+3H2\rightarrow2Fe+3H2O\)
2a 6a 4a
\(\Rightarrow V_{H2}=\left(a+6a\right)\times22.4=9.408l\)
c.nHCl = 0.2 mol
\(Fe+2HCl\rightarrow FeCl2+H2\)
0.1 0.2
m chất rắn còn lại = mCu + m Fe ban đầu - m Fe bị hòa tan
= \(a\times64+4a\times56-0.1\times56=11.68g\)
a) Gọi số mol H2 là x
=> nH2O=x(mol)
Theo ĐLBTKL: mA+mH2=mB+mH2O
=> 200 + 2x = 156 + 18x
=> x = 2,75 (mol)
=> VH2=2,75.22,4=61,6(l)
b) Gọi nCuO=a(mol)
nFe2O3=1,5a(mol)
=> 80a + 240a + 102b = 200
=> 320a + 102b = 200
PTHH: CuO + H2 --to--> Cu + H2O
a---------------->a
Fe2O3 + 3H2 --to--> 2Fe + 3H2O
1,5a------------------>3a
=> 64a + 168a + 102b = 156
=> 232a + 102b = 156
=> a = 0,5; b = \(\dfrac{20}{15}\)
%mCuO=\(\dfrac{0,5.80}{200}\).100%=20%
%mFe2O3=\(\dfrac{0,75.160}{200}\).100%=60%
%mAl2O3=\(\dfrac{\dfrac{20}{15}102}{200}\).100%=20%
c) nH2=\(\dfrac{2,75}{5}\)=0,55(mol)
nFeO(tt)=\(\dfrac{36}{72}\)=0,5(mol)
Gọi số mol FeO phản ứng là t (mol)
PTHH: FeO + H2 --to--> Fe + H2O
t--------------->t
=> 56t + (0,5-t).72 = 29,6
=> t = 0,4 (mol)
=> H%=\(\dfrac{0,4}{0,5}\).100%=80%
Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{Al_2O_3}=2a\left(mol\right)\\n_{CuO}=3a\left(mol\right)\end{matrix}\right.\)
PTHH:
Fe2O3 + 3H2 --to--> 2Fe + 3H2O
a ---------> 3a
CuO + H2 --to--> Cu + H2O
3a ------> 3a
\(\rightarrow3a+3a=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Leftrightarrow a=0,1\left(mol\right)\\ \rightarrow m=0,1.160+0,1.2.102+0,1.3.80=60,4\left(g\right)\)
Gọi \(n_{Fe_2O_3}=x\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=2x\left(mol\right)\\n_{CuO}=3x\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
3x 3x
\(Al_2O_3+3H_2\rightarrow2Al+3H_2O\)
2x 6x
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
x 3x
\(\Rightarrow\Sigma n_{H_2}=3x+6x+3x=0,6\Rightarrow x=0,05mol\)
\(\Rightarrow m=m_{CuO}+m_{Al_2O_3}+m_{Fe_2O_3}\)
\(\Rightarrow m=3\cdot0,05\cdot80+2\cdot0,05\cdot102+0,05\cdot160=30,2g\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe}=a\left(mol\right);n_{Cu}=0,5a\left(mol\right)\\ m_{hhB}=17,6\\ \Leftrightarrow56a+64.0,5a=17,6\\ \Leftrightarrow a=0,2\left(mol\right)\\ \Rightarrow n_{Fe}=0,2\left(mol\right);n_{Cu}=0,1\left(mol\right)\\ a,n_{H_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,2+0,1=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ b,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ \Rightarrow ddC:FeCl_2,HCldư\\ n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)
\(a) n_{Fe_2O_3} = \dfrac{16.75\%}{160} = 0,075(mol)\\ n_{CuO} = \dfrac{16.25\%}{80} = 0,05(mol)\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Fe} = 2n_{Fe_2O_3} = 0,15(mol) \Rightarrow m_{Fe} = 0,15.56 = 8,4(gam)\\ n_{Cu} = n_{CuO} = 0,05(mol) \Rightarrow m_{Cu} = 0,05.64 = 3,2(gam)\\ b) n_{H_2} = 3n_{Fe_2O_3} + n_{CuO} = 0,075.3 + 0,05 = 0,275(mol)\\ V_{H_2} = 0,275.22,4 = 6,16(lít)\)
\(n_{CuO}=2a\left(mol\right)\Rightarrow n_{Fe_2O_3}=a\left(mol\right)\)
\(m_X=80\cdot2a+160a=80\left(g\right)\)
\(\Rightarrow a=0.25\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(n_{H_2}=0.5+0.25\cdot3=1.25\left(mol\right)\)
\(V_{H_2}=1.25\cdot22.4=28\left(l\right)\)
\(m_{cr}=0.5\cdot64+0.5\cdot56=60\left(g\right)\)
\(a,\\ Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)
Loại phản ứng: Phản ứng thế
\(b,n_{Fe}=2.n_{Fe_2O_3}+3.n_{Fe_3O_4}=2.\dfrac{32}{160}+3.0,15=0,85\left(mol\right)\\ m_{Fe}=0,85.56=47,6\left(g\right)\\ c,n_{H_2}=\dfrac{32}{160}.3+4.0,15=1,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,2.22,4=28\left(l\right)\)
Em xem sao oxit sắt lại hỏi KL nhôm nha! Vô lí!!!
