\(CH_4+2O_2\rightarrow CO_2+2H_2O\left(1\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\left(2\right)\)
\(n_{CH_4}=\dfrac{m}{M}=\dfrac{8}{16}=0,5\left(mol\right)\)
Từ \(\left(1\right)\Rightarrow n_{CO_2}=n_{CH_4}=0,5\left(mol\right)\)
\(\left(2\right)\Rightarrow n_{CaCO_3}=n_{CO_2}=0,5\left(mol\right)\)
\(m_{CaCO_3}=n.M=0,5.\left(40+12+3.16\right)=50\left(g\right)\)
Vậy kết tủa \(CaCO_3\) có \(m=50\left(g\right)\)