Đáp án:
\(\begin{array}{l}
a)\\
{m_{{C_2}{H_5}OH}} = 2,5875g\\
b)\\
{m_{C{O_2}}} = 1,65g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_4} + {H_2}O \xrightarrow{t^0,xt} {C_2}{H_5}OH\\
{n_{{C_2}{H_4}}} = \dfrac{{1,68}}{{22,4}} = 0,075\,mol\\
H = 75\% \Rightarrow {n_{{C_2}{H_4}}} \text{tham gia }= 0,075 \times 75\% = 0,05625\,mol\\
{n_{{C_2}{H_5}OH}} = {n_{{C_2}{H_4}}} \text{tham gia }= 0,05625\,mol\\
{m_{{C_2}{H_5}OH}} = 0,05625 \times 46 = 2,5875g\\
b)\\
{n_{{C_2}{H_4}}} \text{dư }= 0,075 - 0,05625 = 0,01875\,mol\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_4}}} \text{dư }= 0,0375\,mol\\
{m_{C{O_2}}} = 0,0375 \times 44 = 1,65g
\end{array}\)
