Đáp án:
Δ = 4,41 gam
Giải thích các bước giải:
\[0,96\ \text{gam}\left\{\begin{matrix} C_2H_2& \\ CH_4 & \\ C_4H_6 & \end{matrix}\right. \xrightarrow{+O_2}\left\{\begin{matrix} CO_2 \xrightarrow{+ 0,05\ \text{mol} \ Ba(OH)_2}\ \downarrow +H_2O& \\ H_2O& \end{matrix}\right.\]
\(n_{Ba(OH)_2}=1\cdot 0,05=0,05\ \text{mol}\)
\(\to n_{\downarrow}=n_{Ba(OH)_2}=0,05 \ \text{mol}\)
\(\to m_{\downarrow}=0,05\cdot 197=9,85\ \text{gam}\)
BT mol C: \(n_{\downarrow}=n_{CO_2}=0,05\ \text{mol}\\\Rightarrow n_C=n_{CO_2}=0,05\ \text{mol}\\\to m_C=0,05\cdot 12=0,6\ \text{gam}\\\to n_{{H\ \text{trong A}}}=0,96-0,6=0,36\ \text{gam}\)
BT mol H: \(n_{H_2O\ \text{đốt A}}=\dfrac{0,36}{2}=0,18 \ \text{mol}\\\to m_{H_2O\ \text{đốt A}}=0,18\cdot 18=3,24\ \text{gam}\)
\(\Rightarrow m_{\text{dd B}}=0,05\cdot 18+3,24=4,14\ \text{gam}\)
\(m_{Ba(OH)_2}=0,05\cdot (137+34)=8,55\ \text{gam}\)
Ta có: \(4,14<8,55\to\)Khối lượng dung dịch B giảm: \(8,55-4,14=4,41\) gam so với dung dịch \(Ba(OH)_2\) ban đầu
