$n_{CH4}=6,72/22,4=0,3mol$
$1/$
$CH4+2O2\overset{t^o}{\rightarrow}CO2+2H2O$
$2/$
theo pt:
$n_{O2}=2.n_{CH4}=2.0,3=0,6mol$
$⇒V_{kk}=0,6.22,4.5=67,2l$
$3/$
$n_{CO2}=n_{CH4}.85\%=0,3.85\%=0,255mol$
$⇒m_{CO2}=0,255.44=11,22g$
$4/$
$CO2+Ca(OH)2→CaCO3↓+H2O$
theo pt:
$n_{CaCO3}=n_{CO2}=0,255mol$
$⇒m_{CaCO3}=0,255.100=25,5g$