Bài $1:$$\begin{cases}n_{CO_2}=\frac{20}{100}=0,2 mol \\ n_{H_2O}=\frac{20-4,9-0,2.44}{18}=0,35 mol\\n_{N_2}=\frac{1,12}{22,4}=0,05 mol \end{cases}$
$*)X:C_xH_yO_z\Rightarrow x:y:z=0,2:(2.0,35):(2.0,05)=2:7:1$
$M_X<70\Rightarrow X:C_2H_7N (CH_3-CH_2-NH_2)$
Bài $2:$
$\begin{cases}n_{CO_2}=\frac{0,672}{22,4}=0,03 mol \\ n_{Ca(OH)_2}=2.0,01=0,02 mol\\n_{NaOH}=\frac{0,4}{40}=0,01 mol \end{cases}$
$\Rightarrow \begin{cases}n_{CO_2}=0,03 mol \\ n_{Ca^{2+}}=0,02 mol\\n_{Na^+}=0,01 mol\\n_{OH^-}=0,05 mol \end{cases}\Rightarrow [\begin{cases}n_{Ca^{2+}}=0,02 mol \\ n_{Na^+}=0,01 mol\\n_{CO_3^{2-}}=0,02 mol\\n_{HCO_3^-}=0,01 mol \end{cases}]\Rightarrow n_{CaCO_3}=0,02 mol\Rightarrow m=2 gam$