Đáp án:
$C_3H_6O$
Giải thích các bước giải:
\(\begin{array}{l}
{n_{Pd}} = \dfrac{{2,12}}{{212}} = 0,01\,mol\\
{n_{CaC{O_3}}} = \frac{3}{{100}} = 0,03\,mol\\
CO + PdC{l_2} \to Pd + C{O_2} + 2HCl(1)\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O(2)\\
{n_{CO}} = {n_{C{O_2}(1)}} = {n_{Pd}} = 0,01\,mol\\
m \text{ bình 1 tăng }= 0,38g \Rightarrow {m_{{H_2}O}} + {m_{CO}} - {m_{C{O_2}(1)}} = 0,38g\\
\Rightarrow {m_{{H_2}O}} = 0,38 + 0,01 \times 44 - 0,01 \times 28 = 0,54g\\
{n_{{H_2}O}} = \dfrac{{0,54}}{{18}} = 0,03\,mol\\
{n_{C{O_2}}} = {n_{C{O_2}(2)}} - {n_{C{O_2}(1)}} = 0,03 - 0,01 = 0,02\,mol\\
CTTQ:{C_x}{H_y}{O_z}\\
{C_x}{H_y}{O_z} + (x + \dfrac{y}{4} - \dfrac{z}{2} - \dfrac{t}{2}){O_2} \to (x - t)C{O_2} + \dfrac{y}{2}{H_2}O + tCO\\
{n_{{H_2}O}} = 0,03\,mol \Rightarrow 0,01 \times \dfrac{y}{2} = 0,03 \Rightarrow y = 6\\
{n_{CO}} = 0,01\,mol \Rightarrow 0,01t = 0,01 \Rightarrow t = 1\\
{n_{C{O_2}}} = 0,02\,mol \Rightarrow 0,01 \times (x - 1) = 0,02 \Rightarrow x = 3\\
{n_{{O_2}}} = \dfrac{{0,784}}{{22,4}} = 0,035\,mol\\
\Rightarrow 0,01 \times (x + \dfrac{y}{4} - \dfrac{z}{2} - \dfrac{t}{2}) = 0,035\\
\Leftrightarrow 0,01 \times (3 + \dfrac{6}{4} - \dfrac{z}{2} - \dfrac{1}{2}) = 0,035 \Rightarrow z = 1\\
CTPT\,X:{C_3}{H_6}O
\end{array}\)
