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Vì: %mCH4 = 80%
\(\Rightarrow m_{CH_4}=25.80\%=20\left(g\right)\Rightarrow n_{CH_4}=\dfrac{20}{16}=1,25\left(mol\right)\)
\(\Rightarrow m_{H_2}=5\left(g\right)\Rightarrow n_{H_2}=\dfrac{5}{2}=2,5\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Theo PT: \(\Sigma n_{O_2}=\dfrac{1}{2}n_{H_2}+2n_{CH_4}=3,75\left(mol\right)\)
\(\Rightarrow V_{O_2}=3,75.22,4=84\left(l\right)\)
Mà: %VO2 = 20%
\(\Rightarrow V_{kk}=\dfrac{84}{20\%}=420\left(l\right)\)
Bạn tham khảo nhé!
a) Gọi số mol Al, Zn là 2a, a (mol)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2a-->1,5a---------->a
2Zn + O2 --to--> 2ZnO
a---->0,5a------->a
=> \(102a+81a=18,3\)
=> a = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)
b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
a, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Cu}=24-m_{Fe}=12,8\left(g\right)\) \(\Rightarrow n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=0,2\left(mol\right)\\n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
⇒ m = mCuO + mFe2O3 = 0,2.80 + 0,1.160 = 32 (g)
b, \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,2.80}{32}.100\%=50\%\\\%m_{Fe_2O_3}=50\%\end{matrix}\right.\)
c, Theo PT: \(n_{H_2}=n_{CuO}+3n_{Fe_2O_3}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)
Bài 1:
a) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+2H_2O\)
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
b) Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20\cdot80\%=16\left(g\right)\\m_{CuO}=20-16=4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,35\left(mol\right)\) \(\Rightarrow V_{H_2}=0,35\cdot22,4=7,84\left(l\right)\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu}=n_{CuO}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{hhB}=m_{Fe}+m_{Cu}=0,2\cdot56+0,05\cdot64=14,4\left(g\right)\)
Bài 2:
PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
a) Vì khối lượng Cu bằng \(\dfrac{6}{5}\) khối lượng Fe
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=\dfrac{26,4}{6+5}\cdot6=14,4\left(g\right)\\m_{Fe}=26,4-14,4=12\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=\dfrac{14,4}{64}=0,225\left(mol\right)\\n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}=\dfrac{9}{28}+0,225=\dfrac{153}{280}\left(mol\right)\) \(\Rightarrow V_{H_2}=\dfrac{153}{280}\cdot22,4=12,24\left(l\right)\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{28}\left(mol\right)\\n_{CuO}=n_{Cu}=0,225\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=\dfrac{3}{28}\cdot160\approx17,14\left(g\right)\\m_{CuO}=0,225\cdot80=18\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=35,14\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{17,14}{35,14}\cdot100\%\approx48,78\%\\\%m_{CuO}=51,22\%\end{matrix}\right.\)