\(n_{O2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(2CO+O2-->2CO2\)(1)
\(2H2+O2-->2H2O\)(2)
\(n_{H2O}=\frac{1,8}{18}=0,1\left(mol\right)\)
\(n_{H2}=n_{H2O}=0,1\left(mol\right)\)
\(\Rightarrow V_{H2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow n_{O2\left(2\right)}=\frac{1}{2}n_{H2O}=0,05\left(mol\right)\)
\(n_{O2\left(1\right)}=0,15-0,05=0,1\left(mol\right)\)
\(n_{CO}=2n_{O2}=0,2\left(mol\right)\)
\(V_{CO}=0,2.22,4=4,48\left(l\right)\)
-tính khối lượng CO2
Cách 1:
\(n_{CO2}=2n_{O2}=0,2\left(mol\right)\)
\(m_{CO2}=0,2.44=8,8\left(g\right)\)
Cách 2
\(m_{O2}=0,1.32=3,2\left(g\right)\)
\(m_{CO}=0,2.28=5,6\left(g\right)\)
Áp dụng ĐLBTKL
\(m_{CO2}=m_{CO}+m_{O2}=3,2+5,6=8,8\left(g\right)\)
