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Đặt \(\left\{{}\begin{matrix}n_{Cu}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol ) \(\Rightarrow n_{hh}=64x+56y=26,4\left(g\right)\) (1)
mà \(64x=1,2.56y\) (2)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,225\\y=\dfrac{3}{14}\end{matrix}\right.\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,225 0,225 ( mol )
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
\(\dfrac{9}{28}\) \(\dfrac{3}{14}\) ( mol )
\(V_{H_2}=22,4.\left(0,225+\dfrac{9}{28}\right)=12,24\left(l\right)\)
\(CuO\left(0,2\right)+H_2\left(0,2\right)\rightarrow Cu\left(0,2\right)+H_2O\)
\(FeO\left(0,3\right)+H_2\left(0,3\right)\rightarrow Fe\left(0,3\right)+H_2O\)
Gọi khối lượng 2 kim loại Fe, Cu lần lược là x, y thì ta có hệ
\(\left\{\begin{matrix}x+y=29,6\\x-y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=16,8\\y=12,8\end{matrix}\right.\)
\(\Rightarrow n_{Fe}=\frac{16,8}{56}=0,3\)
\(n_{Cu}=\frac{12,8}{64}=0,2\)
\(\Rightarrow n_{H_2}=0,3+0,2=0,5\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\)
\(\Rightarrow m=0,2.80+0,3.72=37,6\)
\(m_{Cu}=\dfrac{29,6-4}{2}=12,8(g)\\ \Rightarrow m_{Fe}=12,8+4=16,8(g)\\ PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O\\ Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\ \Rightarrow \Sigma n_{H_2}=n_{Cu}+3n_{Fe}=\dfrac{12,8}{64}+\dfrac{3}{4}.\dfrac{16,8}{56}=0,6(mol)\\ \Rightarrow V_{H_2}=0,6.22,4=13,44(l)\)
Ta có: \(\left\{{}\begin{matrix}m_{Fe}=\dfrac{59,2+8}{2}=33,6\left(g\right)\\m_{Cu}=59,2-33,6=25,6\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\\n_{Cu}=\dfrac{25,6}{64}=0,4\left(mol\right)\end{matrix}\right.\)
PTHH:
\(Fe_3O_4+4H_2\xrightarrow[]{t^o}3Fe+4H_2O\)
0,8<----0,3
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,4<---0,4
`=> V_{H_2} = (0,4 + 0,8).22,4 = 26,88 (l)`
\(CuO\left(0,2\right)+H_2\left(0,2\right)\rightarrow Cu\left(0,2\right)+H_2O\)
\(FeO\left(0,3\right)+H_2\left(0,3\right)\rightarrow Fe\left(0,3\right)+H_2O\)
Gọi khối lượng 2 kim loại Fe, Cu lần lược là x, y thì ta có hệ:
\(\left\{{}\begin{matrix}x+y=29,6\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16,8\\y=12,8\end{matrix}\right.\)
\(\Rightarrow n_{Fe}=\frac{16,8}{56}=0,3\)
\(n_{Cu}=\frac{12,8}{64}=0,2\)
\(\Rightarrow n_{H_2}=0,3+0,2=0,5\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\)
\(\Rightarrow m=0,2.80+0,3.72=37,6\)
Đặt số mol: nPbO=a, nFe2O3=b,pt: PbO+C--->Pb+CO(1),Fe2O3 +3C--->2Fe+3CO(2).theo pt(1) nPb=a mol,theo pt(2) nFe=2.b mol.theo bài ra : a.207+2b.56=26,3 & 207.a=3,696.2b.56 =>a=0,1 &b=0,05.theo pt (1)&pt(2)=>tổng số mol cacbon nC=a+3b=0,25 mol=> mC=0,25.12=3 gam.
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 )
\(m_{CuO}=0,05.80=4g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ Theo.pt:n_{Fe}=n_{H_2}=0,3\left(mol\right)\\ m_{Fe}=0,3.56=16,8\left(g\right)\\ m_{Cu}=20-16,8=3,2\left(g\right)\\ n_{Cu}=\dfrac{3,2}{64}=0,06\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,05\leftarrow0,05\leftarrow0,05\\ m_{CuO}=0,05.80=4\left(g\right)\)
\(\left\{{}\begin{matrix}mCu+mFe=29,6\\mFe-mCu=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mCu=12,8\\mFe=16,8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}nCu=0,2\left(mol\right)\\nFe=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: CuO + H2 -to-> Cu + H2O
0,2____0,2_______0,2(mol)
FeO+H2 -to-> Fe + H2O
0,3___0,3_______0,3(mol)
-> m=mCuO+mFeO= 0,2.80+0,3.72= 37,6(g)
=> m=37,6(g)
V=V(H2,đktc)=(0,2+0,3).22,4=11,2(l)
=>V=11,2(l)