1/
$n_{H_2O}=1,8/18=0,1mol$
$PTHH : 2H_2+O_2\overset{t^o}\to 2H_2O$
$\text{Theo pt :}$
$n_{H_2}=n_{H_2O}=0,1mol⇒V_{H_2}=0,1.22,4=2,24l$
$n_{O_2}=1/2.n_{H_2O}=1/2.0,1=0,05mol⇒V_{O_2}=0,05.22,4=1,12l$
2/
$n_{H_2}=112/22,4=5mol$
$PTHH : 2H_2+O_2\overset{t^o}\to 2H_2O$
$\text{Theo pt :}$
$n_{H_2O}=n_{H_2}=5mol$
$⇒m_{H_2O}=5.18=90g$