a,
PTHH:
$2H_2+O_2\buildrel{{t^o}}\over\to 2H_2O$
b,
$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$n_{O_2}=\dfrac{3,36}{22,4}=0,15(mol)$
$\dfrac{n_{H_2}}{2}<\dfrac{n_{O_2}}{1}$
$\Rightarrow H_2$ hết, $O_2$ dư
$n_{H_2O}=n_{H_2}=0,2(mol)$
$\to m_{H_2O}=0,2.18=3,6g$
$n_{O_2\text{pứ}}=\dfrac{1}{2}n_{H_2}=0,1(mol)$
$\Rightarrow n_{O_2\text{dư}}=0,15-0,1=0,05(mol)$
Vậy khí $B$ là $O_2$ ($0,05$ mol)
c,
$n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$
$3Fe+2O_2\buildrel{{t^o}}\over\to Fe_3O_4$
$\dfrac{0,1}{3}>\dfrac{0,05}{2}$
$\Rightarrow Fe$ dư, $O_2$ hết
$n_{Fe\text{pứ}}=1,5n_{O_2}=0,075(mol)$
$\Rightarrow n_{Fe\text{dư}}=0,1-0,075=0,025(mol)$
$\to m_{Fe\text{dư}}=0,025.56=1,4g$