a, PTHH: Mg+2HCl--->MgCl2+H2
b, nHCl= \(\dfrac{10,95}{36,5}=0,3\) mol
Theo pt: nMg=\(\dfrac{1}{2}.nHCl=\dfrac{1}{2}.0,3=0,15\) mol
=> mMg= 0,15.24= 3,6 (g)
c, Theo pt: nH2=\(\dfrac{1}{2}.nHCl=\dfrac{1}{2}.0,3=0,15\) mol
=> VH2= 0,15.22,4= 3,36 (l)