$a/$
$n_{CH_{4}}=\frac{2,8}{22,4}=0,125mol$
$b/$
$n_{CuO}=\frac{2}{80}=0,025mol$
$c/$
$n_{Cl2}=\frac{1,51.10^{23}}{6.10^{23}}=0,25mol$
$2/$
$a/$
$V_{NH_{3}}=0,25.22,4=5,6l$
$b/$
$n_{SO_{2}}=3,2/64=0,05mol$
$⇒V_{SO_{2}}=0,05.22,4=1,12l$
$c/$
$n_{H_{2}}=\frac{6,02.10^{22}}{6.10^{23}}=0,1mol$
$⇒V_{H_{2}}=0,1.22,4=2,24l.$