Đáp án:
$C_3H_8O_2$
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{m_{{H_2}O}} = m \text{ bình 1 } = 14,4g\\
{n_{{H_2}O}} = \dfrac{{14,4}}{{18}} = 0,8\,mol\\
{m_{CaC{O_3}}} = {m_ \downarrow } = 40g\\
{n_{CaC{O_3}}} = \dfrac{{40}}{{100}} = 0,4\,mol\\
Ca{(HC{O_3})_2} \xrightarrow{t^0} CaC{O_3} + C{O_2} + {H_2}O\\
{n_{Ca{{(HC{O_3})}_2}}} = {n_{CaC{O_3}}} = \dfrac{{10}}{{100}} = 0,1\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} + 2{n_{Ca{{(HC{O_3})}_2}}} = 0,6\,mol\\
{n_C} = {n_{C{O_2}}} = 0,6\,mol\\
{n_H} = 2{n_{{H_2}O}} = 1,6\,mol\\
{m_O} = 15,2 - 0,6 \times 12 - 1,6 = 6,4g\\
{n_O} = \dfrac{{6,4}}{{16}} = 0,4\,mol\\
{n_C}:{n_H}:{n_O} = 0,6:1,6:0,4 = 3:8:2\\
\Rightarrow CTDGN:{C_3}{H_8}{O_2}\\
b)\\
{n_A} = {n_{{N_2}}} = \dfrac{{1,4}}{{28}} = 0,05\,mol\\
{M_A} = \dfrac{{3,8}}{{0,05}} = 76g/mol\\
\Rightarrow 76n = 76 \Rightarrow n = 1\\
CTPT:{C_3}{H_8}{O_2}
\end{array}\)
