Bài 1: nAl=0,1 (mol)
pt: 2Al + 6HCl-> 2AlCl3 + 3H2
vậy:0,1---->0,3----->0,1------>0,15(mol)
b) mAlCl3=n.M=0,1.133,5=13,35(g)
VH2=n.22,4=0,15.22,4=3,36(lít)
c) c) MgO + H2-> Mg + H2O
vậy: 0,15<--0,15-->0,15
=> mMg=n.M=0,15.24=3,6(g)
Bài 2: nFe=0,1 mol
PT: Fe +2 HCl -> FeCl2 + H2
vậy: 0,1--->0,2------>0,1--->0,1 (mol)
b) mFeCl2=n.M=0,1.127=12,7(g)
VH2=n.22,4 =0,1.22,4=2,24(lít)
c) PT: FeO + H2 -> Fe + H2O
vậy: 0,1<----0,1--->0,1(mol)
=> mFe=n.M=0,1.56=5,6(g)