Mg +2HCl\(\rightarrow\)MgCl2+ H2
MgO+2HCl\(\rightarrow\)MgCl2+H2O
ta có nH2 =\(\frac{2,24}{22,4}\)=0,1 mol
Theo pthh 1 :nH2=nMg =0,1 mol
\(\rightarrow\)mMg=0,1.24 =2,4 g
\(\rightarrow\)mMgO =6-2,4=3,6g
nMgO =\(\frac{3,6}{40}\)=0,09 mol
%mMgO =\(\frac{3,6}{6}.100\%\)=60%
Ta co nHCl =2nMg =0,2 mol
nHCl =2nMgO =0,18 mol
mHCl=(0,18+0,2).36,5=13,87 g
VHCl=\(\frac{13,87}{1}.1\)=12,6ml