a, $Zn+H_2SO_4→ZnSO_4+H_2$
$x$ $x$ $mol$
$Fe+H_2SO_4→FeSO_4+H_2$
$y$ $y$ $mol$
$n_{H_2SO_4}=0,25(mol)$
$\left \{ {{x+y=0,25} \atop {65x+56y=14,9}} \right.$
$\left \{ {{x=0,1} \atop {y=0,15}} \right.$
$\text{% m$_{Zn}$=$\frac{0,1·65}{14,9}$·100% = 43,62%}$
$\text{% m$_{Fe}$= 100% − 43,62% =56,38%}$
b, $V_{H_2}=(x+y)·22,4=5,6l$