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\(\left[O\right]_{KL}+H_2->H_2O\\ n_{H_2O}=n_{H_2}=\dfrac{14,4}{18}=0,8mol\\ v=0,8.22,4=17,92L\\ m_{KL}=m=47,2-16.0,8=34,4g\)
a.\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,3 0,3 ( mol )
\(m_{Fe}=0,3.56=16,8g\)
\(\%m_{Fe}=\dfrac{16,8}{20}.100=84\%\)
\(\%m_{Cu}=100\%-84\%=16\%\)
b.\(m_{Cu}=20-16,8=3,2g\)
\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{CuO}=0,05.80=4g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ Theo.pt:n_{Fe}=n_{H_2}=0,3\left(mol\right)\\ m_{Fe}=0,3.56=16,8\left(g\right)\\ m_{Cu}=20-16,8=3,2\left(g\right)\\ n_{Cu}=\dfrac{3,2}{64}=0,06\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,05\leftarrow0,05\leftarrow0,05\\ m_{CuO}=0,05.80=4\left(g\right)\)
n\(H2=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(CuO+H2-->H2O+Cu\)
0,2------0,2(mol)
\(Fe2O3+3H2--->2Fe+3H2O\)
1/15<----0,2(mol)
\(m_{CuO}=0,2.80=16\left(g\right)\)
\(m_{Fe2O3}=\frac{1}{15}.160=\frac{32}{3}\left(g\right)\)
b) \(n_{Cu}=n_{H2}=0,2\left(mol\right)\)
\(m_{Cu}=0,2.64=12,8\left(g\right)\)
\(n_{Fe}=\frac{2}{3}n_{H2}=\frac{2}{15}\left(mol\right)\)
\(m_{Fe}=\frac{2}{15}.56=\frac{112}{15}\left(g\right)\)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)
PTHH: CuO + H2 → Cu + H2O
Mol: x x x
PTHH: Fe2O3 + 3H2 → 2Fe + 3H2O
Mol: y 3y 2y
Ta có hpt:\(\left\{{}\begin{matrix}80x+160y=14\\x+3y=0,225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,075\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(m_{hh.kim.loại}=m_{Cu}+m_{Fe}=0,075.64+2.0,05.56=10,4\left(g\right)\)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)
PTHH:
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo 2 pthh trên: \(n_{H_2O}=n_{H_2}=0,225\left(mol\right)\)
\(\rightarrow m_{H_2O}=0,225.18=4,05\left(g\right)\\ \rightarrow m_{H_2}=0,225.2=0,45\left(g\right)\)
Áp dụng ĐLBTKL, ta có:
\(m_{oxit\left(CuO,Fe_2O_3\right)}+m_{H_2}=m_{\text{kim loại}\left(Cu,Fe\right)}+m_{H_2O}\\ \rightarrow m_{\text{kim loại}\left(Cu,Fe\right)}=14+0,45-4,05=10,4\left(g\right)\)
\(a) n_{Fe_3O_4} = a(mol) ; n_{CuO} = b(mol)\\ \Rightarrow 232a + 80b = 117,6(1)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{H_2} = 4a + b = \dfrac{40,32}{22,4}=1,8(2)\\ (1)(2)\Rightarrow a = 0,3 ;b = 0,6\\ \%m_{Fe_3O_4} = \dfrac{0,3.232}{117,6}.100\% =59,18\%\\ \%m_{CuO} = 100\%-59,18\% = 40,82\%\)
\(b)\\ n_{Fe} = 3a = 0,9(mol)\\ n_{Cu} = b = 0,6(mol)\\ \%m_{Fe} = \dfrac{0,9.56}{0,9.56+0,6.64}.100\% = 56,76\%\\ \%m_{Cu} = 100\% - 56,76\% = 43,24\%\)
\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)
\(m=80a+160b=6\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(n_{H_2}=a+3b=0.1\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.025,b=0.025\)
\(m_{kl}=0.025\cdot64+0.025\cdot2\cdot56=4.4\left(g\right)\)
\(b.\)
\(m_{hh}=3m_{Fe_2O_3}=6\left(g\right)\)
\(\Rightarrow n_{Fe_2O_3}=\dfrac{2}{160}=0.0125\left(mol\right)\)
\(\Rightarrow n_{CuO}=0.0125\left(mol\right)\)
\(m_{kl}=0.0125\cdot2\cdot56+0.0125\cdot64=2.2\left(g\right)\)
nH2 =3,36 /22,4 = 0,15 molH2 + CuO ------> Cu + H2O0,15 0,15=> mCuO = 0,15 . 80= 12g2H2 + Fe2O3 -->2 Fe +3 H2O0,15 0,075 mol=> mFe2O3 = 0,075 . 160 = 12g4H2+Fe3O4→3Fe+4H2O0,15 0,0375 mol=> mfe3O4 = 0,0375 . 232 = 8,7g