Số nguyên tử C trong X:
Vậy X là C3H8
X có dạng CxHy
\(C_xH_y+\left(x+\frac{y}{4}\right)O_2\rightarrow xCO_2+\frac{y}{2}H_2O\)
\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)
\(m_{BaCO3}=29,55\left(g\right)\Rightarrow n_{BaCO3}=\frac{29,55}{197}=0,15\left(mol\right)=n_{CO2}\)
\(\Rightarrow m_{dd\left(giam\right)}=m_{BaCO3}-\left(m_{CO2}+m_{H2O}\right)\)
\(\Rightarrow m_{CO2}+m_{H2O}=29,55-19,35=10,2\left(g\right)\)
\(\Rightarrow m_{H2O}=10,2-0,15.44=3,6\left(g\right)\)
\(\Rightarrow n_{H2O}=\frac{3,6}{18}=0,2\left(mol\right)>n_{CO2}\)
Vậy X là ankan
\(\Rightarrow n_X=n_{H2O}-n_{CO2}=0,2-0,15=0,05\left(mol\right)\)
\(\Rightarrow C_X=\frac{0,15}{0,05}=3\Rightarrow X:C_3H_8\)
