Em tham khảo nha:
\(\begin{array}{l}
7)\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{60}}{{100}} = 0,6\,\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,6}}{2} = 0,3\,mol\\
H = 75\% \Rightarrow {n_{{C_6}{H_{12}}{O_6}}} \text{ cần dùng }= 0,3 \times \dfrac{{100}}{{75}} = 0,4\,mol\\
{m_{{C_6}{H_{12}}{O_6}}} = 0,4 \times 180 = 72g\\
8)\\
{({C_6}{H_{10}}{O_5})_n} + n{H_2}O \xrightarrow{\text{ lên men }} n{C_6}{H_{12}}{O_6}\\
{C_6}{H_{12}}{O_6} \xrightarrow{\text{ lên men }} 2C{O_2} + 2{C_2}{H_5}OH\\
Ca{(HC{O_3})_2} \to CaC{O_3} + C{O_2} + {H_2}O\\
{n_{Ca{{(HC{O_3})}_2}}} = {n_{CaC{O_3}}} = \dfrac{{25}}{{100}} = 0,25\,mol\\
{n_{CaC{O_3}}} = \dfrac{{137,5}}{{100}} = 1,375\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} + 2{n_{Ca{{(HC{O_3})}_2}}} = 1,875\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{1,875}}{2} = 0,9375\,mol\\
H = 81\% \Rightarrow {m_{{{({C_6}{H_{10}}{O_5})}_n}}} \text{ cần dùng }= 0,9375 \times \dfrac{1}{n} \times 162n \times \dfrac{{100}}{{81}} = 187,5g
\end{array}\)
