Hướng dẫn :

${{(-{{C}_{6}}{{H}_{10}}{{O}_{5}}-)}_{n}}+n{{H}_{2}}O\xrightarrow[{{t}^{o}}]{axit}n{{C}_{6}}{{H}_{12}}{{O}_{6}}$

${{C}_{6}}{{H}_{12}}{{O}_{6}}\xrightarrow[ruou,{{t}^{o}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}$

${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{41,4}{46}=0,9$ (mol)

Xét phương trình lên men rượu có :${{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\frac{1}{2}{{n}_{{{C}_{2}}{{H}_{5}}OH}}=\frac{1}{2}.0,9=0,45$ (mol)

Mà hiệu suất $80$%$\Rightarrow {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0,45.0,8=0,36$ (mol)

${{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0,36.180=64,8$ gam

Xét phương trình thủy phân tinh bột:

                    ${{(-{{C}_{6}}{{H}_{10}}{{O}_{5}}-)}_{n}}+n{{H}_{2}}O\xrightarrow[{{t}^{o}}]{axit}n{{C}_{6}}{{H}_{12}}{{O}_{6}}$

Có                162n (g)                          $\leftarrow $      180n (g)

Nên               $\frac{162n.64,8}{180n}$                        $\leftarrow $           64,8 (g)

Mà hiệu suất $80$%$\Rightarrow {{n}_{{{(-{{C}_{6}}{{H}_{10}}{{O}_{5}}-)}_{n}}}}=\frac{162n.64,8}{180n.0,8}=72,9$(g)

                      $\Rightarrow $ Chọn đáp án D.