$PTPƯ:$
$CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O$ $(1)$
$CO_2+Ca(OH)_2\xrightarrow{} CaCO_3↓+H_2O$ $(2)$
$n_{CH_4}=\dfrac{3,2}{16}=0,2mol.$
$Theo$ $pt1:$ $n_{CO_2}=n_{CH_4}=0,2mol.$
$Theo$ $pt2:$ $n_{CaCO_3}=n_{CO_2}=0,2mol.$
$⇒m_{CaCO_3}=0,2.100=20g.$
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