Cho m gam tinh bột men thành C2H5OH với hiệu suất 81%, hấp thụ hết lượng CO2 sinh ra vào dung dịch Ca(OH)2 được 55 gam k

Đáp án A

Phương trình phản ứng:

\(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{O}_{2}}+Ca{{(OH)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O\,\,\,\,\,\,\,\left( 1 \right) \\ & mol\,\,\,\,\,\,\,0,55\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\leftarrow 0,55 \\ \end{align}\) \(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,2C{{O}_{2}}+Ca{{(OH)}_{2}}\to Ca{{(HC{{O}_{3}})}_{2}}\,\,\,\,\,\,\left( 2 \right) \\ & mol\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\leftarrow 0,1 \\ \end{align}\) \(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,Ca{{(HC{{O}_{3}})}_{2}}\xrightarrow{{{t}^{o}}}CaC{{O}_{3}}+C{{O}_{2}}+{{H}_{2}}O\,\,\,\,\,\,\left( 3 \right) \\ & mol\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\leftarrow \,\,\,\,\,\,\,\,\,\,0,1 \\ \end{align}\) \(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,{{C}_{6}}{{H}_{12}}{{O}_{6}}\xrightarrow{le\hat{a}n\,\,men\,\,r\ddot{o}\hat{o}\ddot{i}u}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}\,\,\,\,\,\,\left( 4 \right) \\ & mol\,\,\,\,\,\,0,375\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,75 \\ \end{align}\) \(\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,{{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}+n{{H}_{2}}O\xrightarrow{xt}n{{C}_{6}}{{H}_{12}}{{O}_{6}}\,\,\,\left( 5 \right) \\ & mol\,\,\,\,\,\,\,\,\frac{0,375}{n}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\leftarrow \,\,\,\,\,\,\,\,\,\,\,0,375 \\ \end{align}\)

Theo giả thiết ta thấy khi CO₂ phản ứng với dung dịch Ca(OH)₂ thì tạo ra cả hai loại muối là CaCO₃ và Ca(HCO₃)₂. Từ các phản ứng (1), (2), (3), (4), (5) suy ra:

\({{n}_{{{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}}}=\frac{0,375}{n}mol.\)

Vậy khối lượng tinh bột tham gia phản ứng với hiệu suất 81% là:

\({{m}_{{{C}_{6}}{{H}_{10}}{{O}_{5}}}}=\frac{162n.0,375}{n.81%}=75\,(gam).\)