$PTHH:2H_{2}+O_{2}→2H_{2}O$
$a)n_{H_{2}}=\frac{2,24}{22,4}=0,1mol$
$n_{O_{2}}=\frac{1,68}{22,4}=0,075mol$
Xét tỉ lệ:
$\frac{0,1}{2}<\frac{0,075}{1}⇒O_{2}$ dư
Theo $PTHH:n_{O_{2pư}}=\frac{1}{2}n_{H_{2}}=0,05mol$
⇒$n_{O_{2dư}}=(0,075-0,05).32=0,8 (g)$
$b)$Theo $PTHH: n_{H_{2}O}=n_{H_{2}}=0,1mol$
$⇒m_{H_{2}O}=0,1.18=1,8(g)$