2CO + O2 -> 2CO2 (1)
2H2 + O2 -> 2H2O (2)
nCO2=0,2(mol)
nO2=0,6(mol)
Theo PTHH 1 ta có:
nCO=nCO2=0,2(mol)
nO2(1)=\(\dfrac{1}{2}\)nCO=0,1(mol)
nO2(2)=0,6-0,1=0,5(mol)
Theo PTHH 2 ta có:
nH2=2nO2(2)=1(mol)
mH2=2.1=2(g)
mCO=28.0,2=5,6(g)
%mH2=\(\dfrac{2}{7,6}.100\%=26,3\%\)
%mCO=100-26,3=73,7%
a) PTHH:
2CO + O2 \(\rightarrow\) 2CO2
2H2 + O2 \(\rightarrow2H_2O\)
Ta có: 2CO + O\(_2\)\(\rightarrow\)2CO\(_2\)
2H\(_2\) + O\(_2\)\(\rightarrow\)2H\(_2\)O
N\(_{CO2}\)= 0,2 ( mol )
N\(_{O2}\)= 0,6 ( mol )
Theo PTHH1 ta có:
N\(_{CO}\)= N\(_{CO2}\)= 0,2 ( mol )
N\(_{O2\left(1\right)}\)= \(\dfrac{1}{2}\)N\(_{CO}\)= 0,1 (mol )
N\(_{O2\left(2\right)}\)= 0,6 - 0,1 = 0,5 ( mol )
Theo PTHH2 ta lại có:
N\(_{H2}\)= 2N\(_{O2\left(2\right)}\)= 1 ( mol )
M\(_{H2}\)= 2 . 1 = 2 (g)
M\(_{CO}\)= 28 . 0,2 = 5,6 (g)
%M\(_{H2}\)= \(\dfrac{2}{7,6}.100\%\) = 26,3%
%M\(_{CO}\)= 100 - 26,3 = 73.7 %
Chúc bạn học tốt
b) \(n_{O2}=\dfrac{m_{O2}}{M_{O2}}=\dfrac{9,6}{32}=0,6\left(mol\right)\)
\(n_{CO2}=\dfrac{m_{CO2}}{M_{CO2}}=\dfrac{8,8}{44}=0,2\left(mol\right)\)
theo phương trình hóa học 1 :
\(n_{CO2}=0,2\Rightarrow n_{CO}=0,2\left(mol\right)\)
\(\Rightarrow n_{O2}=0,1\left(mol\right)\)
theo PTHH 2 :
\(\Rightarrow n_{O2}=0,5\left(mol\right)\)
\(\Rightarrow n_{H2}=1\left(mol\right)\)
từ đó :
\(\Rightarrow m_{H2}=M_{H2}.n_{H2}=2.1=2\left(g\right)\)
\(\Rightarrow m_{CO}=M_{CO}.n_{CO}=28.0,2=5,6\left(g\right)\)
\(\Rightarrow\%m_{H2}=\dfrac{2}{7,6}.100=26,3\%\)
\(\Rightarrow\%m_{CO}=100=26,3=73,7\%\)
