Em tham khảo nha:
\(\begin{array}{l}
3)\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
{n_{C{O_2}}} = \dfrac{{35,84}}{{22,4}} = 1,6\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{1,6}}{2} = 0,8\,mol\\
H = 80\% \Rightarrow {n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,8}}{{80\% }} = 1\,mol\\
{C_M}{C_6}{H_{12}}{O_6} = \dfrac{1}{4} = 0,25\,M\\
4)\\
{C_6}{H_{12}}{O_6} {\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{m_{CaC{O_3}}} - {m_{C{O_2}}} = 3,4 \Leftrightarrow {m_{C{O_2}}} = 10 - 3,4 = 6,6g\\
{n_{C{O_2}}} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
H = 90\% \Rightarrow m = {m_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,15}}{2} \times 180 \times \dfrac{{100}}{{90}} = 15g
\end{array}\)
