Thủy phân hoàn toàn tinh bột trong dung dịch axit vô cơ loãng, thu được

Đáp án A

$$\begin{gathered} \begin{array}{l}{\left(C_6{H}_{10}{O}_5\right)}_n+n{H}_{2}O\stackrel{H^{+},t°}{\to }n{C}_6{H}_{12}{O}_6\\ \text{Tinh bột \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}glucozơ}\end{array} \\ \begin{array}{l}C{H}_{2}OH{\left[CHOH\right]}_{4}CHO+H_2\stackrel{Ni,t°}{\to }C{H}_{2}OH{\left[CHOH\right]}_{4}C{H}_{2}OH\\ Sobitol\end{array} \end{gathered}$$