a) PTHH: $CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$ (1)
b) $n_{CH_4}=\dfrac{6,4}{16}=0,4\ mol$
Theo PTHH: $n_{O_2}=2n_{CH_4}=0,8\ mol$
$⇒V_{O_2}=0,8.22,4=17,92\ l$
c) PTHH: $CO_2+Ca(OH)_2→CaCO_3↓+H_2O $ (2)
Theo PTHH (2): $n_{CaCO_3}=n_{CO_2}=0,4\ mol$
$⇒m_{CaCO_3}=0,4.100=40\ g$