Đáp án:
a, \({C_2}{H_4}{O_2}\)
b, \({m_{BaC{O_3}}} = 39,4g\)
c, \(HCOOC{H_3}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{{O_2}}} = 0,2mol\\
{n_{C{O_2}}} = 0,2mol\\
{n_{{H_2}O}} = 0,2mol\\
{M_X} = 60\\
2{n_{{O_2}}} \ne 2{n_{C{O_2}}} + {n_{{H_2}O}}
\end{array}\)
Vì bảo toàn nguyên tố O không bằng nhau nên suy ra trong X có nguyên tố O
\(\begin{array}{l}
\to {n_C} = {n_{C{O_2}}} = 0,2mol\\
\to {n_H} = 2{n_{{H_2}O}} = 0,4mol\\
\to {n_O} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,2mol\\
\to C:H:O = 1:2:1 \to {(C{H_2}O)_n}\\
{M_X} = 60 \to n = 2\\
\to {C_2}{H_4}{O_2}
\end{array}\)
\(\begin{array}{l}
Ba{(OH)_2} + C{O_2} \to BaC{O_3} + {H_2}O\\
\to {n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,2mol\\
\to {m_{BaC{O_3}}} = 39,4g
\end{array}\)
TH1: X là axit \(C{H_3}COOH\)
\(\begin{array}{l}
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{C{H_3}COONa}} = 0,04mol\\
{n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = 0,04mol\\
\to {m_{C{H_3}COOH}} = 2,4g
\end{array}\)
Loại TH1
TH2: X là \(HCOOC{H_3}\)
\(\begin{array}{l}
HCOOC{H_3} + NaOH \to HCOONa + C{H_3}OH\\
{n_{HCOONa}} = 0,05mol\\
{n_{HCOONa}} = {n_{HCOOC{H_3}}} = 0,05mol\\
\to {m_{HCOOC{H_3}}} = 3g
\end{array}\)
