a)0,3 MOL KHÍ CL2
$n_{Cl2}=\frac{m}{M}$ => $m_{Cl2}$= $n_{Cl2}$.M=0,3.(35,5.2)=0,3.71=21,3(g)
b)1,25 MOL H
Tương tự ta cũng có:
$m_{H2}$ =$n_{H2}$ .$M_{H2}$ =1,25.(1.2)=1,25.2=2,5(g)
c)1,12(L)CH4
Ta có:$n_{}=\frac{V}{22,4}$
=> $n_{CH4}$= $\frac{1,12}{22,4}$ =>$n_{CH4}$ =0,05(mol)
Ta có: $n_{CH4}$ =0,05(mol)
=> $m_{CH4}$= $n_{CH4}$ .$M_{CH4}$ =0,05.(12+4)=0,05.16=0,8(g)