$PTPƯ:$
$CO+O\xrightarrow{} CO_2$ $(1)$
$Ca(OH)_2+CO_2\xrightarrow{} CaCO_3+H_2O$ $(2)$
$m_{chất\ rắn\ giảm}=m_{O\ bay\ hơi}=27,6-19,6=8g.$
$⇒n_{O}=\dfrac{8}{16}=0,5mol.$
$Theo$ $pt1:$ $n_{CO_2}=n_{O}=0,5mol.$
$Theo$ $pt2:$ $n_{CaCO_3}=n_{CO_2}=0,5mol.$
$⇒a=m_{CaCO_3}=0,5.100=50g.$
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