$n_{H_2}$ = $\frac{2,8}{22,4}$ = $0,125$ $(mol)$
$2H_2$ + $O_2$ $ →^{t^o}$ $2H_2O$
$0,125$ $0,0625$ $0,125$ $(mol)$
$a.$ $V_{O_2}$ = $0,0625$ . $22,4$ = $1,4$ $( l )$
$m_{O_2}$ = $0,0625$ . $32$ = $2$ $( g )$
$b.$ $m_{H_2O}$ = $0,125$ . $18$ = $2,25$ $( g )$