$n_{H2}=4,8/22,4≈0,214mol$

$2H2+O2\overset{t^o}\to 2H2O$

$a/$

Theo pt :

$n_{H2O}=n_{H2}=0,214mol$

$⇒m_{H2O}=0,214.18=3,852g$

$b/$

$n_{O_2}=1/2.n_{H2}=1/2.0,214=0,107mol$

$⇒V_{O_2}=0,107.22,4=2,3968l$

$2/$

$n_{hh}=n_{CO2}+n_{O2}+n_{H2}=\dfrac{22}{44}+\dfrac{1,6}{32}+\dfrac{0,2}{2}=0,5+0,05+0,05=0,6mol$

\(\begin{array}{l}
1)\\
a)\\
2{H_2} + {O_2} \to 2{H_2}O\\
n{H_2} = \dfrac{{4,8}}{{22,4}} = \dfrac{3}{{14}}\,mol\\
n{H_2}O = n{H_2} = \dfrac{3}{{14}}mol\\
m{H_2}O = \dfrac{3}{{14}} \times 18 = 3,86g\\
b)\\
n{O_2} = \dfrac{3}{{28}}\,mol\\
V{O_2} = \dfrac{3}{{28}} \times 22,4 = 2,4l\\
2)\\
nC{O_2} = \dfrac{{22}}{{44}} = 0,5\,mol\\
n{O_2} = \dfrac{{1,6}}{{32}} = 0,05\,mol\\
n{H_2} = \dfrac{{0,2}}{2} = 0,1\,mol\\
M = \dfrac{{22 + 1,6 + 0,2}}{{0,5 + 0,1 + 0,05}} = 36,6g/mol
\end{array}\)