PTHH : \(2H_2+CO_2\underrightarrow{_{t^o}}CH_4+2H_2O\)
\(n_{CO_2}=\dfrac{m}{M}=\dfrac{8,8}{44}=0,2\left(mol\right)\)
Bài ra : 2mol-----1mol------1mol-----2mol
Suy ra: 0,2mol-----0,1mol------0,1mol-----0,2mol (*)
\(\Rightarrow V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow V_{CO_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)
Theo tỉ lệ số mol (*) ta có :
\(m_{H_2}=n.M=0,2.2=0,4\left(g\right)\)
\(m_{CO_2}=n.M=0,1.44=4,4\left(g\right)\)
\(\Rightarrow m_{hh}=0,4+4,4=4,8\left(g\right)\)
nCO2=\(\dfrac{8,8}{44}=0,2mol\)
2H2+O2->2H2O (1)
2CO+O2->2CO2 (2)
do chỉ có PT( 2 ) sinh ra CO2
=>VCO=0,2.22,4=4,48 lít
=>VH2=22,4-4,48=17,92 lít
b,=>VO2 PT (1)=\(\dfrac{17,92}{2}=8,96l\)
=>VO2 ở PT (2)=\(\dfrac{4,48}{2}=2,24l\)
=>V cần dùng là 8,96+2,24=11,2 lít
c, mCO=\(\dfrac{4,48}{22,4}\times28=5,6g\)
mH2=\(\dfrac{17,92}{22,4}\times2=1,6g\)
m hỗn hợp =5,6+1,6=7,2g
d, VCo2=0,2.22,4=4,48l