a,
m giảm= mO= 13,38-12,456= 0,924g
H2+ O -> H2O
=> nH2= nO= 0,924/16= 0,05775 mol
V H2= 0,05775.22,4= 1,2936l
b,
PbO+ H2 -> Pb+ H2O
Gọi x là mol Pb trong A, y là mol PbO dư
=> 207x+ 223y= 12,456 (1)
nPbO pu= nPb= x mol
=> 223x+ 223y= 13,38 (2)
(1)(2)=> x= 0,05775; y= 0,0025
%Pb= $\frac{0,05775.207.100}{12,456}$= 95,97%
%PbO= 4,03%