\(n_{BaCO_3}=\dfrac{29.55}{197}=0.15\left(mol\right)\)
\(n_{CO_2}=n_{BaCO_3}=0.15\left(mol\right)\)
\(m_{giảm}=m_{BaCO_3}-\left(m_{CO_2}+m_{H_2O}\right)=19.35\left(g\right)\)
\(\Rightarrow m_{H_2O}=29.55-19.35-0.15\cdot44=3.6\left(g\right)\)
\(n_{H_2O}=\dfrac{3.6}{18}=0.2\left(mol\right)\)
\(n_{H_2O}>n_{CO_2}\Rightarrow Xlà:ankan\)
\(CT:C_nH_{2n+2}\)
\(\dfrac{n}{2n+2}=\dfrac{0.15}{0.2\cdot2}\Rightarrow n=3\)
\(CT:C_3H_8\)
