Đáp án:
4 B
5 B
6 D
Giải thích các bước giải:
\(\begin{array}{l}
4)\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{360}}{{180}} = 2\,mol\\
{n_{C{O_2}}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 4\,mol\\
H = 80\% \Rightarrow {n_{C{O_2}}} \text{ thu được }= 4 \times 80\% = 3,2\,mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 3,2\,mol\\
{m_{CaC{O_3}}} = 3,2 \times 100 = 320g\\
5)\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{50}}{{100}} = 0,5\,\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,5}}{2} = 0,25\,mol\\
H = 80\% \Rightarrow {n_{{C_6}{H_{12}}{O_6}}} \text{ cần dùng } = 0,25 \times \dfrac{{100}}{{80}} = 0,3125\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = 0,3125 \times 180 = 56,25g\\
6)\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{80}}{{100}} = 0,8\,\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,8}}{2} = 0,4\,mol\\
H = 75\% \Rightarrow {n_{{C_6}{H_{12}}{O_6}}} \text{ cần dùng } = 0,4 \times \dfrac{{100}}{{75}} = \dfrac{8}{{15}}\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{8}{{15}} \times 180 = 96g
\end{array}\)
