a, 2H2 + O2 → 2H2O
b, $n_{H2}$ =$\frac{6.72}{22.4}$ =0.3 (mol)
$n_{02}$ = $\frac{n_{H2}}{2}$ = $\frac{0.3}{2}$ = 0.15 (mol)
=> $m_{O2}$ = 0.15 x 16 = 2.4 (g)
$V_{O2}$ = 0.15 x 22.4 = 3.36 (L)
c, $n_{H2}$ = $n_{H2O}$ = 0.3 (mol)
=> $m_{H2O}$ = 0.3 x 18 = 5.4 (g)