a,Ta co pthh
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
Theo de bai ta co
nFe=\(\dfrac{5,6}{56}=0,1mol\)
Theo pthh
nH2=nFe=0,1mol
\(\Rightarrow\)VH2 (dktc)=0,1.22,4=2,24 l
b, Theo pthh
nHCl=2nFe=2.0,1=0,2mol
\(\Rightarrow\)mct=mHCl=0,2.36,5=7,3 g
\(\Rightarrow mddHCl=\dfrac{mct.100\%}{C\%}=\dfrac{7,3.100\%}{30\%}\approx24,33g\)
c, Theo pthh
nFeCl2=nFe=0,1 mol
\(\Rightarrow\) mFeCl2=0,1.127=12,7 g